exercise 5.1: trigonometric ratios - kopykitab · exercise 5.1: trigonometric ratios 1.) find the...

Exercise 5.1: Trigonometric Ratios

1.) Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given.

Sol.

(i)sinA=23sinA — |

Given:

sinA=23sin.A = | .....(1)

By definition,

SinA=23sin^4 = | = P e r p e n d i c u l a r H y p o t e n u s e ....(2)

By Comparing (1) and (2)

We get,

Perpendicular side = 2 and

Hypotenuse = 3

Therefore, by Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side

(AB)

Therefore,

32 = AB2 + 22

AB2 = 32 - 22

AB2 = 9 - 4

AB2 = 5

AB= V5\/5

Hence, Base = \5 \ /5

Now, COSA= BaseHypotenuseCOS A = Hyf™„use

COSA= \53 COS A =

Now, cosec A = isinA -r—r

Therefore,

Hypotenusecosec A = HypotenusePerependicular -=------ -77— 7—Perependicu lar

cosec A =321

HypotenuseNow, sec A = HypotenuseBase----Base-----

Therefore,

sec A = 3V5 -^=

Now, tan A = PerependicularBase PereP^^i ular

tan A = 2a/5

Now, cot A = BasePerpendicular -=-------T.— —Perpendicular

Therefore,

(ii) COsA—45cos A — |

Given: COsA=45cos^4. = .... (1)

By Definition,

COSA=BaseHypotenuseCOS A = „ Ba:se----- .... (2)Hypotenuse '

By comparing (1) and (2)

We get,

Base =4 and

Hypotenuse = 5

B A

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side

(BC)

52 = 42+ BC2

BC2 = 52 - 42

BC2= 2 5 -1 6

BC2 = 9

BC= 3

Hence, Perpendicular side = 3

Now,

■ * • « O , c / u c iS in A = 2 3 s m A = 3 = PerpendicularHypotenuse Hypotenuse

Perpendicular

Therefore,

sinA=35sinA = |

Now, cosec A = 1sinA -t—t-sinA

Therefore,

cosec A= 1 sinA —sinA

Therefore,

cosec A =HypotenusePerependicularHypotenuse

Perependicular

c o s e c A = 53

Now, sec A = 1 cosA — t’ cosA

Therefore,

sec A = HypotenuseBaseHypotenuse

Base

sec A = 54

. . . » _ „ PerpendicularNow, tan A = PerpendicularBase-----Base--------

Therefore,

tan A =34 44

Now, cot A = 1tanA — —7tanA

Therefore,

cot A = BasePerpendicular BasePerpendicular

COt A = 43 3

(iii) tan0=llltan© = y-

Given: ta n 0 = n ita n © = -y .... (1)

By definition,

. ^ ^ Perpendiculart a n 0 = PerpendicularBasetanH ------ .... (2)


We get,

Base= 1 and

Perpendicular side= 5

Therefore,


AC2 = AB2 + BC2

Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse(AC)

AC2 = 12 + 112

AC2 = 1 + 121

AC2= 122

AC= V l2 2 v /122

. , ■ PerpendicularNow, SIR ©— PerpendicularHypotenuse Sin 0 = 77— 7— —

Therefore,

sin0 =iiVi2 2 sin© = ^ =

Now, cosec0=lsinO@ —

cosec 0=V12211 0 =

Now, COS©= BaseHypotenuse cos 0 BaseHypotenuse

Therefore,

cosO= 1 \T22cos 0 = -^=

Now, secO=icososec 0 = —C O S 0

Therefore,

sec0= HypotenuseBasesec0 = sec©=V1221 sec 0 = sec©=Vl22

sec © = Vl22

Now, COtO=1tan0cot 0 =

Therefore,

COt0= BasePerpendicularCOt 0 — — — Ba']e , COtO=11lCOt© —Perpendicular l l

(iv) sin0=iH5sin© — ^

Given: s in © = lH 5 s in © = .... (1)

By definition,

Sin©— PerpendicularHypotenuse sin © PerpendicularHypotenuse . . . (2 )


We get,

Perpendicular Side = 11 and

Hypotenuse= 15

Therefore,


AC2 = AB2 + BC2

Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

152 = AB2 +112

AB2 = 152 - 112

AB2= 225-121

AB2 = 104

AB = Vl04\/l04

AB= V2x2x2x13\/2 x 2 x 2 x 13

AB= 2V2x13\/2 x 13

AB= 2V26\/26

Hence, Base = 2^26<\/26

Now, COS0= BaseHypotenuse COS © BaseHypotenuse

Therefore,

COS©= 2V2615 COS @ =

Now, cosec[©= ls in © 0 =s in 0

Therefore,

cosec©— HypotenusePerpendicular©Hypotenuse

Perpendicular

c o s e c © = 1 5 1 1 0 = - j j

HypotenuseNow, sec©— HypotenuseBase © — -----Base------

Therefore,

s e c © = 152V26 0 = ~ ^ =2y 26

Now, ta n © - PerpendicularBasetan 0 PerpendicularBase

Therefore,

tan©=ii2\26tan0 =

Now, COt0— BasePerpendicularCOt 0 = BasePerpendicular

Therefore,

COt©=2V2611 cot 0 = —qrj—■

(v)tana=5i2tana = ^

Given: tana=5l2ta im = ^ .... (1)

By definition,

tana— PerpendicularBase t a n OtPerpendicular

Base (2)


We get,

Base= 12 and

Perpendicular side = 5

Therefore,


AC2 = AB2 + BC2

Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC)

AC2 = 122 + 52

AC2 = 144 + 25

AC2= 169

AC= 13

Hence Hypotenuse = 13

Now, Sina- PerpendicularHypotenusesin a = PerpendicularHypotenuse

Therefore,

s in a= 5 i3 s in a = ^

Now, COSecQ- HypotenusePerpendicularQ!Hypotenuse

Perpendicular

c o s e c a = i 3 5 a — -jr-

Now, COSQ— BaseHypotenuse COS Qt BaseHypotenuse

Therefore,

cosa=i2i3cosa — ^

Now, seca=icosaseca = —cosa

Therefore,

COtQ= BasePerpendicularcot O t = - = —Bas,e , C O ta = 1 2 5 c o ta = ^P e rv e n d ic u la r o

(vi) sin0=V 32sin© = ^

Given: sin©=V32sin@ = ^ .... (1)

By definition,

. . r\ Perpendicular / 0 .SI n 0 — P erpendicularHypotenuse S in 0 = Hypotenuse


We get,

Perpendicular Side = V 3 \/3

Hypotenuse = 2

Therefore,


AC2 = AB2 + BC2


22 = AB2 + (V3\/3)2

AB2 = 22 - (V3-s/3)2

AB2 = 4 - 3

AB2 = 1

AB= 1

Hence Base = 1

Now, COSO= BaseHypotenuse COS 0 = HypZnuse

Therefore,

COS©= 12COS 0 =

Now, COSecG=1sin0© — -r^ r

Therefore,

cosec©- HypotenusePerpendicualar© =Hypotenuse

Perpendicualar

Now, See©- HypotenuseBase sec 0Hypotenuse

Base

Therefore,

sec©=2 i sec© = j-

Now, ta n © - PerpendicularBasetan 0 = PerpendicularBase

Therefore,

tan0=V3i ta n © = ^

Now, COt0 — BasePerpendicularCOt 0 = BasePerpendicular

Therefore,

(vii) C O S © — 725COS © — ^T

Given: COS0=725cos@ = .... (1)

By definition,

COS©= BaseHypotenuse COS 0 „ B + StH y p o t e r


We get,

Base = 7 and

Hypotenuse = 25

Therefore


AC2= AB2 + BC2

Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side(BC)

252 = 72 +BC2

BC2 = 252 - 72

BC2 = 625 - 49

BC = 576

BC= V576a/576


Now, S iflO - perpendicularHypotenusesi.il 0 perpendicularHypotenuse

Therefore,

s in © = 242 5 s in 0 = ^

Now, cosecO=lsin©0 —sin 0

Therefore,

CO SecO - HypotenusePerpendicualar 0Hypotenuse

Perpendicualar

COSeC0= 2524 0 = | |

Now, secO=icos©sec0 = —^cos 0

Therefore,

S e C 0 = HypotenuseBasesec © — HVPotenus_e_ s e c @ = 2 5 7 sec 0 _

. , , ^ l PerpendicularNow, tan0- PerpendicularBasetan 0 = ---- ~Base--------

Therefore,

tan0=247ta n © — -y-

Now, COt0= 1tan0 cot 0 = - ^ 7 7tan 0

Therefore,

COt©= BasePerpendicularCOt 0 -=—Ba e~,— COt0= 724COt0Perpendicular

(viii) t a n 0 = 8 i 5 t a n 0 = ^

Given: tan©=8l5tan0 = .... (1)

By definition,

tanO- PerpendicularBasetan 0 = PerPe™dicul,ar (2 )


We get,

Base= 15 and


Therefore,


AC2= 152 + 82

AC2= 225 + 64

AC2 = 289

AC = a/289^/289

AC= 17

Hence, Hypotenuse = 17

Now, S in0— PerpendicularHypotenusesin 0 PerpendicularHypotenuse

Therefore,

S in0=8l7sin@ = ^

Now, cosecO= ls in © 0 — -7 —-s in 0

Therefore,

COSecO— HypotenusePerpendicular 0Hypotenuse

Perpendicular

Now, COS0— BaseHypotenuse COS 0 = Hyp^Zuse

Therefore,

COSO= 1517 COS 0 — Yf

Now, Sec0=1cos0sec© = —cos 0

Therefore,

SeC0= HypotenuseBasesec0 — SeC0= 1715sec 0 —

Now, COtO= 1 tan© cot 0 = —’ tan©

Therefore,

C O t0= BasePerpendicularCOt 0 — —— Ba™ ,— C O tO =158C O t0Perpendicular

(ix) C O tO — 125 COt 0 = - y

Given: CO t0=i25cot© = ^ ■■■■ (1)

By definition,

COt©= 1 tan© cot 0 — — ^rtan ©

COt0= BasePerpendicularCOt 0 = -5— Ba*f ,— .... (2)


We get,

Base = 12 and


Therefore,


AC2= AB2 + BC2

Substituting the value of base side (AB) and perpendicular side(BC) and get the hypotenuse (AC)

AC2 = 122 + 52

AC2= 144 + 25

AC2 = 169

AC = V i 6 9 \/ l6 9

AC = 13

Hence, Hypotenuse = 13

Now, Sin©— PerpendicularHypotenusesin © PerpendicularHypotenuse

Therefore,

s in © = 5 i3 s in 0 = Tjj

Now, cosec©= ls in 0 @ =s in 0

Therefore,

COSecO— HypotenusePerpendicular©Hypotenuse

Perpendicular

cosec© = 1350 =

Now, C O S 0— BaseHypotenuse COS 0 BaseHypotenuse

Therefore,

COSO= 1213COS 0 — -jl

Now, secO= ic o s o s e c 0 =

Therefore,

SecO= HypotenuseBase sec 0 = SecO = 1312sec @ —

Now, tanO= 1 cot© ta n 0 = —cot ©

Therefore,

tanO=PerpendicularBasetan© = PerPe™dicular fa n © — 512ta n ©

(x) sec©=i35sec@ — ^

Given: sec0=l35sec@ = (1)

By definition,

sec©= i c o s o s e c © = — .... (2)


We get,

Base=5

Hypotenuse = 13

Therefore,


132 = 52 + BC2

BC2 = 132- 5 2

Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side(BC)

BC2=169 - 25

BC2= 144

BC= V l4 4 \ / i4 4

BC = 12


Now, Sin©— PerpendicularHypotenusesin 0 PerpendicularHypotenuse

Therefore,

s in © = i2 i3 s in © —

Now, cosec O = ls in © 0 =sin 0

Therefore,

C O S e c O — HypotenusePerpendicular©Hypotenuse

Perpendicular

c o s e c © = 1312 0 = j |

Now, COS0= 1 sec© cos 0 = —sec 0

Therefore,

COS0= BaseHypotenuse COS 0 = „ Bcf e---- COS0=513COS0 = ^Hypotenuse 13

Now, ta n 0 - PerpendicularBasetan 0 PerpendicularBase

Therefore,

tan 0 = 1 25 ta n © = 0

1tan ©

Now, COtO- 1tan0cot 0 =

Therefore,

COtO- BasePerpendicularCOt © — — — Base — COtO— 512 c o t 0 — -prPerpendicular IZ

Given:

(xi) cosecO=VlO© =

cosec0=Vioi0 = (1)

By definition

COSecO=1sin0@ = —r—77 ....(2)sin 0 ' '

0 — HypotenusePerpendicular© HypotenusePerpendicular

By comparing (1) and(2)

We get,

Perpendicular side= 1 and

Hypotenuse = V i OV^HT

Therefore,


AC2 = AB2 + BC2


(Vi 0 \/I0 )2 = a b 2 + 12

AB2= (V10x/1 0 )2 - 1 2

a b 2= 10 -1

AB = V9\/9

AB = 3

Hence, Base side = 3

Now, S iflO - PerpendicularHypotenuseSm ©Perpendicular

Hypotenuse

Therefore,

s in 0 = iV io s in © 1v/lo

Now, COS0— BaseHypotenuse COS © BaseHypotenuse

Therefore,

cos©= 3vTo cos 0 = - =■v/10

Now, sec©= icosOsec 0 = —cos 0

Therefore,

SeC0= HypotenuseBase sec 0 = SeC0=WlO3sec

. , . /-v ± PerpendicularNow, ta n 0 - PerpendicularBasetan 0 = ----------------

Therefore,

tan 0 = 13 ta n 0 =

Now, C O t O = Ita n O C O t 0 =

cot©= 31 c o t 0 = | cot0=3cot 0 = 3

( x i i ) C O S© 1215 c o s 0 ^ |

Given: C O S © 1 2 1 5 C O S 0 | | .... (1)

By definition,

COS0 — BaseHypotenuseCOS 0 = Hyp^ nuse ■■■■ (2 )


We get,

Base=12 and

Hypotenuse = 15

Therefore,


AC2 = AB2+ BC2

Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

152 = 122 + BC2

BC2 = 152 - 122

BC2 = 225 - 144

BC 2= 81

bc = V81 V sT

BC = 9


Now, SinO-PerpendicularHypotenusesin ©Perpendicular

Hypotenuse

Therefore,

sin0=9i5sin© _9_15

Therefore,

Now, cosec©= isin© © = -sin 0

cosec ©-HypotenusePerpendicular©Hypotenuse

Perpendicular

cosec©= 1 5 9 0 = y

Now, sec©=icososec© =

Therefore,

SeC0= HypotenuseBasesec0 = SeC0= 1512sec0 = |§

Now, ta n © - PerpendicularBasetan 0 = PerPe™dicular

Therefore,

tan©=9i2tan0 — y

Now, COt©= ItanOCOt 0 = —^ r’ tan ©

Therefore,

COt©= BasePerpendicularCOt © = —— Bas,e , C O t0=129C O t© = ^Perpendicular y

2.) In a AAABC, right angled at B , AB - 24 cm , BC= 7 cm , Determine

(i) sin A , cos A

(ii) sin C, cos C

Sol.

(i) The given triangle is below:

c

A

Given: In A A ABC , AB= 24 cm

BC = 7cm

Z A B C Z A B C = 90°

To find: sin A, cos A

In this problem, Hypotenuse side is unknown

Hence we first find hypotenuse side by Pythagoras theorem


We get,

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = V625\/625

AC= 25

Hypotenuse = 25

By definition,

S in A — PerpendicularsideoppositetozAHypotenusesin A = Hypotenuse

sinA=BCACsinil = sinA=725sin A 7_25

By definition,

C O S A - BasesideadjacenttozAHypotenuse COS A =

c o s A = 41 C O S A = 2425 COS A — | |

BasesideadjacenttoZAHypotenuse c o s A = a bac

Answer:

sinA= 725sin A = cosA= 2425cos A = | |

(ii) The given triangle is below:

Given: In AA ABC , AB= 24 cm

BC = 7cm

zABCZABC = 90°

To find: sin C, cos C

In this problem, Hypotenuse side is unknown

Hence we first find hypotenuse side by Pythagoras theorem


We get,

AC2 = AB2 + BC2

AC2 = 242 + 72

AC2 = 576 + 49

AC2= 625

AC = V625\/625

AC= 25

Hypotenuse = 25

By definition,

SinC— PerpendicularsideoppositetozCHypotenuse s in C —

sinC=a ba c s in C = sinC= 2425 s in C = U

Hypotenuse

By definition,

COSC— BasesideadjacenttozCHypotenuseCOS C

c o s A — COSA= 725 COS A — 4=r

BasesideadjacenttoZC COSA= BCACHypotenuse

Answer:

sinA= 2425sin A = C O SA =725C O sA = ^

3.) In the below figure, find tan P and cot R. Is tan P = cot R?

In the given right angled A P Q R A PQR, length of side OR is unknown

Therefore , by applying Pythagoras theorem in A P Q R A PQR

We get,

PR2 = PQ2 + QR2

Substituting the length of given side PR and PQ in the above equation

132= 122 + QR2

p

To find, tan P, cot R

Sol.

QR2 = 132 - 122

q r 2 = 169-144

QR2= 25

QR = V25v/25

By definiton, we know tha t,

tanP- PerpendicularsideoppositetozPBasesideadjacenttozP tan P

tanP=QRPQtan P =

PerpendicularsideoppositetoZPBasesideadjacenttoZP

ta n P = 5 i2 ta n P = .... (1)

Also, by definition, we know that

COtR— BasesideadjacenttozRPerpendicularsideoppositetozR

cot R — Base side adjacent to ZR Perpendicular side opposite to ZR

COtR=QRPQcot R QRPQ

co tR = 5 i2co tR = ^ .... (2)

Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.

Therefore, L.H.S of both equations is also equal

tan P = cot R

Answer:

Yes , tan P =cot R = 512 ^

4.) If sin A = 941 Compute cos A and tan A.

Sol.

Given: s inA =94 ls inA = .... (1)

To find: cos A, tan A

By definition,

SinA— PerpendicularsideoppositetozAHypotenuse sin A Perpendicular side opposite to ZA Hypotenuse


We ge t,


Hypotenuse = 41

Now using the perpendicular side and hypotenuse we can construct AABCAABC as shown below

Length of side AB is unknown is right angled AABCAABC ,

To find the length of side AB, we use Pythagoras theorem,

Therefore, by applying Pythagoras theorem in AABCAABC ,

We get,

AC2 = AB2 + BC2

412 = AB2 + 92

AB2 = 412 - 92

AB2 = 168-81

AB= 1600

AB = V1600^/1600

AB = 40

Hence, length of side AB= 40

Now

By definition,

COSA= BasesideadjacenttozAHypotenuse COS A

COS A = COSA= 4041 COS A =

Base side adjacent to ZA A------------rT— t---------------- COSA= ABACHypotenuse

Now,

By definition,

tanA— PerpendicularsideoppositetozABasesideadjacenttozA

ta n A = SECSl tanA=BCABtanA = -fg tanA=940tanA = £Base s*ae adjacent to Z A A B 4U

Answer:

C O S A = 4041 cos A = , tanA =940tan A — ^

5.) Given 15cot A=8, find sin A and sec A.

Answer:

Given: 15cot A = 8

To find: sin A, sec A

Since 15 cot A =8

By taking 15 on R.H.S

We get,

COtA=8i5cot A = ^

By definition,

COtA= ItanACOt A — -^—r

Hence,

cotA= 1 PerpendicularsideoppositetozABasesideadjacenttozA cot A __________1_________Perpendicular side opposite to ZA

Base side adjacent to ZA

COtA— BasesideadjacenttozAPerpendicularsideoppositetozA

Base side adjacent to Z Acot A —

Perpendicular side opposite to Z A

Comparing equation (1) and (2)

We get,

Base side adjacent to Z A Z A = 8

Perpendicular side opposite to Z A Z A = 15

A A B C A A B C can be drawn below using above information

Hypotenuse side is unknown.

Therefore, we find side AC of AABCAABC by Pythagoras theorem.

So, by applying Pythagoras theorem to AABCA A B C

We get,

AC2 = AB2 +BC2

Substituting values of sides from the above figure

AC2 = 82 + 152

AC2 = 64 + 225

AC2 = 289

AC = V289>/289

AC = 17

Therefore, hypotenuse =17

Now by definition,

S in A — PerpendicularsideoppositetozAHypotenuse s in APerpendicular side opposite to / .A

Hypotenuse

Therefore, sinA=BCACsin A — ^’ A C


sinA=i5i7sinA =

By definition,

secA=icosAsec = -A _cos A

Hence,

secA= 1 BasesideadjacenttozAHypotenuse S e C A ________ 1________Base side adjacent to Z.A

Hypotenuse

SecA— HypotenuseBasesideadjacenttozA se c A — HypotenuseBase side adjacent to / .A


secA=i78sec A = ^

Answer:

sinA=i5i7sinz4 — secA=i78secz4 — ^

6.) In AP QRAPQR , right angled at Q, PQ = 4cm and RQ= 3 cm .Find the valueof sin P, sin R , sec P and sec R.

Sol.

Given:

A P Q R A PQR is right angled at vertex Q.

PQ= 4cm

RQ= 3cm

To find,

sin P, sin R , sec P , sec R

Given APQRAPQR is as shown below

p

<1

Hypotenuse side PR is unknown.

Therefore, we find side PR of APQRA PQR by Pythagoras theorem

By applying Pythagoras theorem to APQRA PQR

We get,

PR2 = PQ2 +RQ2


PR2 = 42 +32

PR2 = 25

PR = V25V25

PR = 5

Hence, Hypotenuse =5

Now by definition,


sinP=35sinP — |

PR2 = 16 + 9

SinP — PerpendicularsideoppositetozPHypotenuse s in P =

SinP = RQPRsin P =

Hypotenuse

Now by definition,

S in R — PerpendicularsideoppositetozRHypotenuse s in RPerpendicular side opposite to Z R

Hypotenuse

sinR=PQPRsin# —

Substituting the values of sides from above figure

sin R= 45 sin# = 40By definition,

secP = icosP se c P = ----------- S e C P — 1 BasesideadiacenttozpHypotenuse SeC P — —5------ r;— y . ------ 7 Z — /—1 r 1 r Base side adjacent to ZipHypotenuse

S e c P — HypotenuseBasesideadjacenttozP se c P — HypotenuseBase side adjacent to Z P


secP = prpqsec P = ^ secP= 5 4 sec P = f

By definition,

secR=icosRsec# 1cos R

secR= 1BasesideadjacenttozRHypotenuse S e C R ______1______Base side adjacent to ZR

Hypotenuse

SecR— HypotenuseBasesideadjacenttozRsec R HypotenuseBase side adjacent to Z R


secR=PRRQsec# — secR =53sec# = |

Answer:

sinP =35sinP = | , sinR=45sin# =

secP=54sec P = secR=53sec# = |

7.) If cot0=78cot @ = | , evaluate

(i) 1+sin0xl-s in0 1+cos0x1-cos0 1+sin Q x 1-sin 0 1+cos 0 x 1-cos 0

(ii) COt20cot2 0

Sol.

Given: C O t© = 7 8 c o t© —

To evaluate: l+sin0xl-sin0l+cos0xl-cos0 1+ sin® x sin®l+ co s 0 x l-cos 0

. . _ . . _ . _ 1+sin 0 x 1-sin 0 ,l+ co s 0 x 1-cos 0 ' '

We know the following formula

(a + b)(a - b) = a2 - b2

By applying the above formula in the numerator of equation (1)

We get,

(1 +sin0)x(1-sin0)=1-sin20.... (2)(Where,a=1 andb=sin0)(1 + sinO) x (1 -sinff) = 1 -sin29........ (2) (Where, a = 1 and b = sin0)

Similarly,

By applying formula (a +b) (a - b) = a2 - b2 in the denominator of equation (1).

We get,

(1+cos0)(1— cos0)=12—cos20 ( l + cos@ )(l-cos@ ) = l 2-cos2 0 ... (Where a=1

and b= COS0cos 0

(1+COS0)(1—COS0)=1—COS20 (1 + c o s 0 )( l-c o s 0 ) — 1-cos2 0 ... (Where a=1 and

b= cosOcos 0

Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).

Therefore,

(1 +sin0)(1-sin0)(1 +cos0)(1-cos0) (1+sm^ ^ ~ Sm^ = 1-sin201-cos20 1-sin ® ....(4 )' A A A ' (l+ cos 0 ) (1-cos 0 ) l-co s2 0 v ’

Since,

cos20+sin20=1cos2 © + sin2 0 = 1

Therefore,

cos20=1 -s in2 0cos2 0 = 1-sin2 0

Also, sin20 = 1 -cos20sin2 0 — 1-cos2 0

Putting the value of 1— sin2© 1 - sin2 © and 1— COS201-co s2 © in equation (4)

We get,

(l+sin 0 )(l-sin 0 ) „ , cos2 0(1+sin0 )(1-sin0 )(1+cos0 )(1-cos0 ) 77-------777-:-------= cos20 sin20 — 5—(1+COS0) (1-cos 0 ) sin2 0

We know that, cos0sin0=COt©-er L^- — cot©sin 0

(1 +sin0)(1-sin0)(1 +cos0)(1-cos0) q ) = ( C o t 0 )2 ( c o t 0 ) 2

Since, it is given that COt©=78cot 0 = |

Therefore,

(1 +sin0)(1-sin0)(1 +cos0)(1-cos0)(l+sin 0 ) (1-sin 0 ) (1+cos 0 ) (1-cos 0 ) = (78)2( | ) 2

(1 +sin0)(1-sin0)(1 +cos0)(1-cos0)(l+sin 0 ) (1-sin 0 ) _ 2. 2 ^(1+cos 0 ) (1-cos 0 ) 82

(1 +sin0)(1-sin0)(1 +cos0)(1-cos0)(l+ sin 0 ) (1-sin 0 )

(1+cos 0 ) (1-cos 0 ) = 49641

(ii) Given: COt©=78cot 0 = |

To evaluate: COt20co t2 ©

COt0= 78 COt © — |

Squaring on both sides,

We get,

(cot©)2=(78)2(c o t0 )2 = ( | ) 2

(COt©)2(cot 0 ) 2 =4964 | |

Answer:

4949641

1—tan2 A _ COg2 s jn 2 ^ Qr nQj 1+tan2 A

Sol.

Given: 3cot A =4

To check whether i-tan2Ai +tan2A=COS2A—sin2A 1~tan*\At — cos2 A -s in 2 A or not.l+ ta n 2 A

8.) If 3cotA=43 cot A — 4 , check whether i-tan2Ai+tan2A=COS2A—sin2A

3cot A =4

Dividing by 3 on both sides,

We get,

cot A = 431 .... (1)

By definition,

COtA= ItanACOt A — —tan A

Therefore,

COtA= 1 PerpendicularsideoppositetozABasesideadjacenttozA COt A = Perpendicular opposite to ZA~Base side adjacent to Z A

CO tA— BasesideadjacenttozAPerpendicularsideoppositetozA

Base side adjacent to Z Acot A —Perpendicular side opposite to / .A -■ (2 )

Comparing (1) and (2)

We get,

Base side adjacent to Z A Z A = 4

Perpendicular side opposite to z A Z A = 3

Hence AABCAABC is as shown in figure below

In AABCAABC , Hypotenuse is unknown

Hence, it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem inAABCAABC

We get

AC2= AB2 +BC2

Substituting the values of sides from the above figure

AC2 =42 + 32

AC2 = 16 +9

AC2 = 25

AC = V25v/25

AC = 5

Hence, hypotenuse= 5

To check whether

To check whether l-tan2Al +tan2A=COS2A—sin2A }~tWL\A. = cos2 A -s in 2 A or not.l+ tan2 A

We get thee values of tan A , cos A , sin A

By definition,

tan A = icotA—cot A

Substituting the value of cot A from equation (1)

We get,

tan A = 14 -j4

tan A = 3 4 4 .... (3)

Now by definition,

* a Base side adjacent to ZA ACOSA— BasesideadjacenttozAHypotenuse COS A = --------TT ^___ ____ COSA— ABACHypotenuse

COS A = # AC

Substituting the values of sides from the above figure

COSA=45COS A = | .... (5)

Now we first take L.H.S of equation l-tan2Al+tan2A=COS2A—sin2A

l W A = c o s 2 ^ _ sip2 ^1+tan2 A

L.H.S = 1-tan2A1+tan2A 1-tan2 A 1+tan2 A

Substituting value of tan A from equation (3)

We get,

L.H.S= 1- ( 34)21+(34)2H f ) 2i+ ( l) 2

l - f - l21-tan2A1 +tan2A 1 tan2^ = 1—(34)21 +(34 )2 -----

1+tan2 A v ' v ' l + ( - ) 2

j __ 9_

1-tan2A1+tan2A } tan,^ = 1—9161+916 161+tan2 A 1 + —^ 16

Taking L.C.M on both numerator and denominator

We get,

1-tan2A1 +tan2A } tan2„^ 1+tan2 A

= 16-91616+916

16-916

16+916

725 ■■■■ (8)

Now we take R.H.S of equation whether 1-tan2A1+tan2A=COS2A—sin2A

= cos2 A -s in 2 A1+tan2 A

R.H.S = cos2A-sin2Acos2 A - sin2 A

Substituting value of sin A and cos A from equation (4) and (5)

We get,

R . H . S = ( 4 5 ) 2- ( 3 5 2 ) ( f ) 2 - ( f 2 )

cos2A-sin2Acos2 A - sin2 A= (45)2-(352) ( | ) 2- ( | 2)

cos2A-sin2Acos2 A - sin2 A = 16 25- 9 2 5 1 |-

cos2A-sin2Acos2 A - sin2 A = 16 -9 25

cos2A-sin2Acos2 A-sin2 A = 7 2 5 ^ ....(7)

Comparing (6) and (7)

We get.

1-tan2A1 +tan2A=cos2A-sin2A }~tanl AA = cos2 A-sin2 Al+ ta n 2 A

Answer:

Yes, 1 -tan2Ai +tan2A =cos2 A—s i n2 A A. — cos2 A-sin2 Al+ ta n 2 A

9.) If tanO =abtan@ = f0 find the value of cosO+sinOcosO-sinO cos Q+sin Q cos 0 -s in 0

Sol.

Given:

tan© =abtan© — | .... (1)

Now, we know that tan©=sin0cos0tan ©

Therefore equation (1) become as follows

. _ _ sin© _ . asinOcos©----- 77 - ab -rcos 0 0

Now, by applying invertendo

We get,

. cos 0 b cosOsinO — ba . „ — — sin 0 a

Now by applying Componendo - dividendo

We get,

cos0+sin0cos0-sin0 — b+ab-acos 0 + s in 0 cos 0 -s in 0

b+ab-a

Therefore,

cosO+sinOcosO-sinO = b+ab-a cos 0 + s in 0 cos 0 -s in 0

b+ab-a

10.) If 3tanO=43 tan 0 = 4 , find the value of 4cosO-sin02cosO+sinO

Sol.

Given: If 3tan©=43 tan 0 = 4

Therefore,

tan© =43tan© = | .... (1)

Now, we know that tan©=sin0cos0tan 0 —cos 0

Therefore equation (1) becomes

sinQcosQ = 43 sin® 4 ■■■■(2)cos 0 3 ' '

Now, by applying Invertendo to equation (2)

We get,

cosQsinQ = 34 cos® = 4 .... (3) sin 0 4 v '

Now, multiplying by 4 on both sides

We get

4x cosOsinO =4X34 4 X ^ 7 7 = 4 X 7sin 0 4

Therefore

4cos0-sin0sin0 = 3-11 4 0 -s in 0 = 3=1sm 0 1

4 cos 0 -s in 0 2 cos 0 + s in 0

4cos©-sin©sin© = 21 4cos.e ^ in 0 = | .... (4)sin © 1 v '

Now, multiplying by 2 on both sides of equation (3)

We get,

2cos0sin0 = 32 2cos0 = f sin 0 2

Now by applying componendo in above equation

o 0 0 2cos 0 + s i n 0 3+22cos0+sin0sin0 - 3 + 2 2 -------- :——--------- = —F—sin 0 2

„ ^ ^ ^ 2 cos 0 + s in 0 52cos0+sin0sin0 — 5 2 -------t ~ ^ ------- = — ....(5 )sin 0 2 ' '

We get,

4cos©-sin0sin0 2cos0+sin0sin0 — 21 52

4 c o s 0 - s i n 0

s in 6

2 c o s 0 + s i n 0

s in 6

21_5

2

Therefore,

4cos0-sin0sin0 x sin02cos0+sin0 = 21 x 25 4 cos e ^ m 0sin 0

X sin 02 cos 0 + s in 0

21

Therefore, on L.H.S s inO sin© cancels and we get,

4cos0-sin02cos0+sin0 = 21 x 25 4 cos0~sin 0 = 7 X f2 cos 0 + s m 0 1 5

Therefore,

4cosO—sinO=44 c o s 0 - s in 0 = 4

11.) If 3cotO=23 cot 0 = 2, find the value of 4sinO-3cosQ2sinO+6cosO

Sol.

Given:

3cotO=23 cot 0 = 2

Therefore,

C O t O = 2 3 c o t 0 = | . . . . (1 )

25

4 sin 0 - 3 cos 0 2 sin 0 + 6 cos 0

Now, we know that COt0=cos0sin0cot © — sin ©


cos0sin0 = 23 c?s® ....(2 )sin © o v 7

Now , by applying invertendo to equation (2)

sin0cos0 = 3 2 | ....(3)cos © 2, v 7

Now, multiplying by 43 4 on both sides,O

We get,

43 x sin0cos0 = 43 x 32 4 X = 4 X |3 cos 0 3 2

Therefore, 3 cancels out on R.H.S and

We get,

. . _.0 4 sin© 24sin03cos0 — 21 ------ rr T3 cos 0 1

Now by applying invertendo dividendo in above equation

We get,

. . _ 0 „ . . 4 sin 0 - 3 cos 0 2-14sin0-3cos03cos0 — 2 -1 1 ----- ------ =------ — —3 cos © 1

„ . _ 0 _ _ AA 4 sin 0-3 cos 0 1 tA\4sin0-3cos03cos0 — 11----- ~ ^ ------ = -r . . . .(4)3 c o s 0 1 ' '

Now, multiplying by 261 on both sides of equation (3)

We get,

26 x sin0cos0 = 26 x 32 -§ X = | X |6 cos 0 6 2


We get,

„ . ______ 2 sin 0 3 „ . ^ „ 2 sin 0 12sin06cos0 — 36 ----- tt -t 2sin06cos0 — 12 ----- -t6 cos 0 b 6 cos 0 2

Now by applying componendo in above equation

We get,

2cos0+6sin06sin0 — 1 +222 cos 0 + 6 sin 0

6 sin 01+2

2

2cos0+6sin06sin0 — 32 2 cos 0 + 6 sin 0 6 sin 0

32 ' ...(5)

Now, by dividing equation (4) by (5)

We get,

4 sin 0 - 3 c o s 0 1

_ 3 s in 0 ____ 14sin0-3cos03sin02cos0+6sin06sin0 — 11 32 - r ------~ c .—pr- — ~~T

2 c o s 0 + 6 s in 0 £

6 s in 0 2

Therefore,

4sin0-3cos03sin0 * 6sin02cos0+6sin0 = 11 * 23 4 sin e 73 c°s 0 x - — 6-fin 0 . _ = \3 sin 0 2 cos 0 + 6 sin 0 1

4sin0-3cos03sin0 * 2x3sin02cos0+6sin0 = 11 * 23 4 sm® 73 ™ s 0 X 2x^ ^ 0 =3 sin 0 2 cos 0 + 6 sin 0

Therefore, on L.H.S (3 Sin0sin0) cancels out and we get,

„ , . ^ ^ ^ „ v „ „ 2 x 4 sin 0 - 3 cos 0 1 w 22x4sin0-3cos02cos0+6sin0 — 11 x 23 —-----_ , _ . _ = 7 X 72 cos 0 + 6 sm 0 1 3

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

. . 0 „ 4 sin 0 -3 cos 0 24sin©-3cos02cos0+6sin0 — 23x2 -------_ , _ . _ = ^ —772 c o s 0 + 6 s m 0 3x2


We get,

. . 0 _ . 0 4 sin 0 - 3 cos 0 14sin©-3cos02cos0+6sin0 —13 7;------„ , „ . _ ■=■2 cos 0 + 6 sm 0 3

Hence answer,

. . _ . _ . _ . _ _ . _ 4 sin 0 - 3 cos 0 14sinO-3cos02cosO+6sinO —13 ------ „ , _ . _ — -5-2 cos 0 + 6 sin 0 3

12.) If tanO = abtan0 = ^ , prove that asin0-bcos0 asin0 +bcos0 = a2-b2a2+b2

a sin 0 - 6 cos 0 __ a2-6 2a sin 0 + 6 cos 0 a2+ 6 2

Sol.

Given:

ta n O a b ta n © ! .... (1)

Now, we know that tan©=sin0cos0tan © =cos 0


sin0cos0 = ab sin® ^ ■■■■(2)COS 0 0 v 7

Now, by multiplying by ab | on both sides of equation (2)

We get,

abx sin0cos0 = ab Xabf- X ^ 7 7 = x f-6 cos 0 o o

Therefore,

asin0bcos0 — a2b2 ° sin® ....(3)6 cos 0 b2

Now by applying dividendo in above equation (3)

We get,

asin0-beos0bcos0 — a2-b2b2 asm'9 b5.osQ = a ■■■■(4)o cos© b2.

Now by applying componendo in equation (3)

We get,

asin0+bcos0bcos0 — a2+b2b2a sin 0 + 6 cos 0 6 cos 0

a2+6262

....(5)

Now, by dividing equation (4) by equation (5)

We get,

asin©-bcos0 bcos0 asin0 +bcos0 bcos0 — a2-b 2b2 a2+b2b2

a s in 0 - b c o s Q

b c o s Q

a s in Q + b c o s Q

b c o s ©

c?-&ft2

a 2 + 6 2

ft2

Therefore,

asin0-bcos0bcos0 x bcos0asin0+bcos0 = a2-b2b2 x b2a2+b2

a sin 0 - 6 cos 0 v 6 cos 0 __ a2—b2 626 cos 0 a sin 0 + 6 cos 0 62 a2 + 6 2

Therefore, bcosObcos © and b2 cancels on L.H.S and R.H.S respectively

asin©-bcos0asin0+bcos0 — a2-b 2a2+b2 a sin 0 -6 cos Q a sin 0 + 6 cos 0

a2—l

a2+62

Hence, it is proved that

asin0 -bcos0 asin0 +bcos0 = a2-b 2a2+b2 a sin 0 -6 cos 0 a sin 0 + 6 cos 0

a2—62 g2+62

13.) If sec0=135sec@ = show that 2sinQ-3cosQ4sinQ-9cosQ =3 l cos ®7 5 4 sin 0 - 9 cos 0

Sol.

Given:

sec0=i35sec@ — 5

To show that 2sin0-3cos04sin0-9cos0 =3 \ s|n ® j? cos ® = 34 sin 0 - 9 cos 0

Now, we know that COS©= lsec0cos © — — ^rsec 0

Therefore,

COS©= 1135 COS 0 = -jj-T

Therefore,

COS0=513c o s@ = ^ .... (1)

Now, we know that

C O S 0 — Basesideadjacenttoz0Hypotenuse COS 0Base side adjacent to ZO

Hypotenuse

Now, by comparing equation (1) and(2)

We get,

Base side adjacent to Z 0 Z © = 5

And

Hypotenuse =13

Therefore from above figure

Base side BC = 5

Hypotenuse AC = 13

Side AB is unknown. It can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

AC2 = AB2 + BC2

Therefore by substituting the values of known sides

We get,

132 = AB2 + 52

Therefore,

AB2= 132 - 52

AB2= 169-25

AB2 = 144

AB = V144VT44

Therefore,

AB = 12 .... (3)

Now, we know that

s in© = abac s in © =A C

s in © = i2 i3 s in © = .... (4)

Now L.H.S of the equation to be proved is as follows

L.H.S = 2sinQ-3tanQ4sinQ-3cosQ \ sln ® 1*tan ®4 sin 0 - 3 cos 0

Substituting the value COS©cos 0 of sinOsin 0and from equation (1) and (4) respectively

We get,

2x 1213— 3X5134* 1213— 9*5132x — -3 x — 13 6 x 13

4x — -9 x —13 y X 13

Therefore,

L.H.S = 2 x 1 2 - 3 x 6 4 x 1 2 - 9 x 5

L.H.S = 24-1548-45

L.H.S= 931

L.H.S= 3

Hence proved that,

_ . _ _ . . _ _ _ 2 sin0 - 3 ta n © _ 02sinO-3tan04sinO-3cosO . . _ „---- rr- = 34 sin 0 - 3 cos 0

14.) If c o s 0 = 1 2 1 3 c o s 0 = , show that sinO(1-tanO)=35i56

sin 0 ( 1 - ta n 0 ) — ^

Sol.

Given: COS0=1213COS0 = | | .... (1)

To show that S in0(1—tan0)=35l56sin 0 (1 —ta n 0 ) =

Now we know that COS0= BasesideadjacenttozGHypotenuse COS 0 =

....(2)

Therefore, by comparing equation (1) and (2)

We get,

Base side adjacent to Z 0 Hypotenuse

Base side adjacent to Z 0 Z 0 = 12

And

Hypotenuse = 13

Therefore from above figure

Base side BC= 12

Hypotenuse AC= 13

Side AB is unknown and it can be determined by using Pythagoras theorem


We get,

AC2= AB2 + BC2


We get,

132= AB2 + 122

Therefore,

AB2= 132- 122

AB2= 169-144

AB =25

AB= V25\/25

Therefore,

AB = 5 .... (3)

Now, we know that

................. . Perpendicular side opposite to Z 0SinW- PerpendicularsideoppositetozOHypotenuseSin fc) = ---------- jj—~i------------------------

Now from figure (a)

sin©=abac s in © —A C

Therefore,

S in© =5 l2s in0 = ^ .... (5)

Now L.H.S of the equation to be proved is as follows

L.H.S of the equation to be proved is as follows

L.H.S = s in0(1-tan0]s in@ (l-tan@ ] .... (6)

Substituting the value of sinOsin 0 and tanOtan 0 from equation (4) and (5)

We get,

L.H.S = 513(1-512)^(1 -

Taking L.C.M inside the bracket

We get,

L-H.S= 513(lx121«12 -512 )^(ifif -

Therefore,

L.H.S= 513(12-512)1| ( ^ - )

L.H.S= 5 1 3 (7 1 2 )^ (^ )

Now by opening the bracket and simplifying

We get,

L.H.S= 5 x 7 1 3 x 1 2 ^

L.H.S= 35136^

From equation (6) and (7) ,it can be shown that

that sin0(1—tan0)sin 0 ( 1 - ta n 0 ) = 35136^

15.) If C O t© =lV 3C O t 0 = -4= , show that 1-cos202-sin20 = 35 1~cos„ ® f’ 2-sin2 0 5

We get,

Sol.

Given: COt©= W3cot @ ..(1)

To show that 1-cos202-sin20 = 35 1 cos2 0 2-sin2 0

35

N o w , w e k n o w that COtO= 1 t a n 0 c o t 0 - —tan 0

Since ta n © - Perpendicularsideoppositetoz0Basesideadjacenttoz0

tan© = Perpendicular side opposite to ZQ Base side adjacent to ZQ

Therefore,

ta n © = i PerpendicularsideoppositetozOBasesideadjacenttozQ tan© ________ l________Perpendicular side opposite to Z Q

Base side adjacent to ZQ

Therefore,

C O t0—Basesideadjacenttoz.0Perpendicularsideoppositetozi0

Base side adjacent to ZQCOt O Perpendicular side opposite to ZQ

Comparing Equation (1) and (2)

We get.

Base side adjacent to Z .0 Z 0 = 1

Perpendicular side opposite to Z.QZQ =

Therefore, triangle representing angle V 3\/3 is as shown below

Therefore, by substituting the values of known sides

We get,

AC2 = (V3)2(-s/3 )2 + 12

Therefore,

AC2 = 3+1

AC2= 4

AC= V 4 \/4

Therefore,

AC = 2 .... (3)

Now, we know that

„■ „ ................. . Perpendicular side opposite to Z 0S in W - PerpendicularsideoppositetozOHypotenuseSin fc) = -------------- jj—~i------------------------r Hypotenuse

Now from figure (a)

S i n 0 = A B A C s i n © = AC

Therefore from figure (a) and equation (3),

s in©=\32sin 0 = ^

Now we know that

C O S 0 BasesideadjacenttozOHypotenuse COS 0Base side adjacent to Z 0

Hypotenuse

Now from figure (a)

We get,

BCBCAC AC

Therefore from figure (a) and equation (3),

C O S 0 = 12COS 0 = \ . . . . (5)

Now, L.H.S of the equation to be proved is as follows

L.H.S = 1-cos2Q2-sin2Q 1 cos„ 02-sin2 0

Substituting the value of from equation (4) and (5)

We get,

Now by taking L.C.M in numerator as well as denominator

We get,

( 4 x l ) - l

L.H.S= (4x1H 4(4x2)-34

4

Therefore,

4-1L.H.S = 4-148-34 -^g-

4

L.H.S = 34x 45 X |

L.H.S = 35 4 = R.H.S0

Therefore,

A ■ 2 n - o r 1-COS2 0 31-cosz02-sinz0 — 3 5 --------5— -=-2 -s in 2 0 5

16.) lftan0=iV7tan0 = then show that cosec20 -sec20 cosec20 +sec20

cosec2 0 -se c 2 0 _ ^ jl cosec2 0 + sec2 0 4

Sol.

Given: tan©=lV7tan@ = ~ = .... (1)

To show that cosec20 -sec2Ocosec20+sec20 cosec ® s e c ® = 34 4cosed10+sec^ 0 4

Now, we know that

Since, tan©— PerpendicularsideopositetozOBasesideadjacenttozO

tail 0 _ Perpendicular side oposite to Z 0Base side adjacent to Z 0

Therefore,

Comparing equation (1) and (2)

We get.

Perpendicular side opposite to Z0Z@ = 1

Base side adjacent to Z0Z@ = V7y/7

Therefore, Triangle representing Z0Z@ is shown below

Hypotenuse AC is unknown and it can be found by using Pythagoras theorem


We get,

AC2= AB2 + BC2


We get,

AC2= (1)2 + (V7)2( \ /7 )2

Therefore,

AC 2 = 1 +7

AC2 = 8

AC = y l S V 8

AC = V2x2x2V2 x 2 x 2

Therefore,

AC = 2V22\/2 .... (3)

Now we know that

SinO-PerpendicularsideoppositetozOHypotenuse s in 0

sin0=a ba c s in 0 —

Perpendicular side opposite to Z 0 Hypotenuse

sinO = i2V 2sin0 = .... (4)

Now, we know that cosec 0 = isin© 0 lsin 0

Therefore, from equation (4)

We get,

cosec 0 = 2 ^ 2 0 = 2y/2 .... (5)

Now, we know that

COS©= BasesideadjacenttozOHypotenuse COS 0Base side adjacent to Z 0

Hypotenuse

Now from figure (a)

We get,

CO S©=BCACCO S© = ^ 7

Therefore from figure (a) and equation (3)

C O S© =V72V2CO S0 = .... (6 )

Now we know that sec0= 1 cos© sec 0 = —cos 0

Therefore, from equation (6)

We get,

sec©= 1V72V2 sec 0 =2s/2

sec©= 2\2\7s e c 0 — .... (7)

Now, L.H.S of the equation to be proved is as follows

L.H.S cosec20 -sec20cosec20+sec20 cosec2 0 -se c 2 0 cosec2 Q + sec2 0

Substituting the value of cosec0© andsecOsec 0 from equation (6) and (7)

We get,

2 2 — 2 2 [(V2)]1‘-{^4)'L.H.S = [(2V2)] -(2V2V7) [(2V2)] +(2V5V7) v

[(2V2)]S+ ( 2 f )

L.H.S= (8 ) _ ( 87 ) (8 )+ ( 87 )w - d )

(8>+d)

Therefore,

56-8756+87

5 6 -8

7

5 6 + 8

48

L.H.S = 487647 -g j-

Therefore,

L.H.S = 4864 o4

L.H.S = 34 4 = R.H.S

Therefore,

cosec20 -sec20cosec20+sec20 cosec2 0-sec2 0 _ cosec2 0+sec2 0

Hence proved that

cosec20-sec20cosec20+sec20 cosec0^ ~ sec0f .cosec2 B+sec2 B

17.) If sec0=54sec 0 = |,find the value of sin0-2cos0tan0-cot0

Sol.

Given: sec0=54sec0 = -| .... (1)

To find the value of sin0-2cos0tan0-cot0 sm ® 2 cos ®tan 0 -c o t 0

Now we know that sec©= icos0sec 0 cos 0

sin 0 -2 cos 0 tan 0 -cot 0

Therefore,

C O S O = 1sec0 COS 0 —

Therefore from equation (1)

COSO= 15COS 0 =

C O S0=45C O S@ — | .... (2)

Also, we know that COS2O+sin20=1cos2 0 + sin2 0 = 1

Therefore,

sin20 = 1 -cos2Osin2 0 = 1-cos2 0 s in0=V l—cos2Osin© — V I-c o s 2 0

Substituting the value of COS0cos ©from equation (2)

We get,

Therefore,

s inO =35sin0 = | .... (3)

Also, we know that

sec20=1 +tan20sec2 0 = 1 + ta n 2 0

Therefore,

Therefore,

ta n 0 = 3 4 ta n 0 = | .... (4)

Also, COt©= ItanOcot 0 1tan 0


We get,

COt©=43COt© = | .... (5)

Substituting the value of COS0cos 0 , COtOcot 0 and tanOtan© from the equation (2),(3),(4) and (5) respectively in the expression below

sin0-2cos0tan0-cot0 sin 0 -2 cos 0 tan 0 -cot 0

We get,

I — 2(±)sin0-2cos0tan0-cot0 S*D ® 2 cos ® = 35- 2 (45)3 4 -4 3 —j —tan 0 -cot 0 ' ' £ _ i

Therefore, sinQ-2cosQtanQ-cotQ sn ® 2 cos ® = 127 4tan 0 -cot 0 7

18.) If S in 0 = l2 l3 s in 0 = , find the value of 2sinQcosQcos2Q-sin2Q 2 slon ® cos®cos2 0 -s in ©

Sol.

Given: S in 0 = l2 l3 s in 0 = j | .... (1)

To, find the value of 2sin0cos0cos20-sin20 2 si„n ® cos ®cos2 0-sin2 0

Now, we know the following trigonometric identity

cosec2 0=1 +tan2© 0 = 1 + ta n 2 0

Therefore, by substituting the value of tanO tan 0 from equation (1)

We get,

2 2cosec2 0 = 1 + ( 12 13 ) 0 = 1 + ( ^ | )

= 1 + 122132 1 + ± 4 132

= 1 + 1441691 + | | |

By taking L.C.M on the R.H.S

We get,

cosec20 = 169+1441690 = 1691~t| 44lb9

= 313169 313169

Therefore

COSeC0=V313169 0 = y j H I

= 0 = V 3 1 3 1 3 0 =

Therefore

cosecO© = 0= V313130 = .... (2)13

Now, we know that

cosec0cosec0 = isino^p-s in fc)

sin0=iV3i3i3sin0 = -^=-13

Therefore

s in0= i3V 3 i3s in0 — .... (3)y 313


cos20+sin20=1cos2 © + sin2 0 = 1

Therefore,

cos20=1— sin20cos2 0 = 1-s in2 0

Now by substituting the value of Sin0sin0 from equation (3)

We get,

2 2 COS20 = 1 —( 13V313 ) cos2 0 = 1- ^ - j = )

= 1 -1 6 9 3 1 3 1 -m

Therefore, by taking L.C.M on R.H.S

We get,

COS2© = 144313 COS2 0 = | | |

Now, by taking square root on both sides

We get,

C O S © = 12V313 COS 0 = —=V313

Therefore,

COS0= 12V313 COS 0 = - j = .... (4)

Substituting the value of sinQsin @ and COS0COS 0 from equation (3) and (4) respectively in the equation below

o ■ 1 r \ ■ 2 r \ 2 sin © C O S 02sin0 cos0 cos20 -sin 0 — — ---- 5—cos2 0-sin 0

Therefore,

2sin0COS0COS20 - S in 20 2 sin ® COS ® = 2X13V313*12V313(13V313)2-(12V313)2cos2 0-sin 0 (

2x 13 12----- , V313

13

312_ 313- 31231325313 - j j "

313

3 1 2 2 531225

Therefore

2sinOcosOcos20-sin20 2 sin ® cos ® =cos2 0-sin 0

31225 31225

19.) If COsO— 35 cos 0 = | , find the value of sinO-itano2tanOsin 0 -

ta n 0

2 tan 0

Sol.

Given: COS0=35cos0 — | .... (1)

To find the value of sin0-itano2tan0 — —2 tan ©

Now we know the following trigonometric identity

cos20 + s in 20=1cos2 0 + sin2 0 = 1

Therefore by substituting the value of COS0COS 0 from equation (1)

We get,

(35)2+sin20=1 ( | ) 2 + sin2 0 = 1

Therefore,

sin20 = 1 -(3 5 )2sin2 0 = 1 - ( | ) 2 sin20=1-(925)s in2 0 = l - ( J r ) s in20=25-925

sin2 0 = sin20 = 1625 sin2 0 = -||

Therefore by taking square root on both sides

We get,

S in0= 45s in0 = | .... (2)

Now, we know that

tan0=sin0cosOtan0 = cos ©

Therefore by substituting the value of S in 0 s in 0 and COS0cos0 from equation (2) and (1) respectively

We get,

4

tanO=4535 = 4 3 ta n 0 = -§- = | .... (4)7

Now, by substituting the value of S in0sin 0 and of ta n 0 ta n © from equation (2) and equation (4) respectively in the expression below

sin 9 - —^sin0-itane2tan0 — ----- ^ —2 tan©

We get,

1t a n 8

= 4 5 -1 4 2 X 4 3sin 0 -

sinO-itano2tan0 2 tan 0

sinO-itanG2tan0 sin 0 ~ ^ ¥ 2 tan 0 = 1620-152083

4 15 4

| CO

XCN

16 15

20 20

sin0-itane2tan0sin 0 ——^_________ tan 8

2 tan 0 = 3160 160

Therefore,

sin0-itano2tanOsin0 tanq _ . . . 3

2 tan© " J 160

cos 0 -20.) If s inO =35sin0 = | , find the value of cos0-itanG2coto —- — rp re

' 5 2 cot 0

Sol.

Given:

s in0=35s in© = | .... (1)O

cos 0 - — ^To find the value of cos0-itane2cot0 — - — —2 cot 0


cos20 + s in 2O=1cos2 0 + sin2 0 = 1

Therefore by substituting the value of COS0cos 0 from equation (1)

We get,

cos20+ (35)2=1cos2 © + ( | ) 2 = 1

Therefore,

COS20 = 1 —(35)2COS2 © = 1 - ( | ) 2 COS20 = 1-925COS2 0 = 1 - ^ -

Now by taking L.C.M

We get,

COS20 = 25-925 COS2 0 = COS20 = 25-925 COS2 © = ^jjr-

Therefore, by taking square roots on both sides

COS©— 45 COS 0 — |

Therefore,

COS©= 45COS 0 = | .... (2)

Now we know that

tan© — sinOcosOtan © = sm®cos 0

Therefore by substituting the value of s inQ sin© and COS0cos© from equation (1) and (2) respectively

We get,

_3

tan©= 3545 tan. 0 = -7-_45

tan© =34tan© = | .... (3)

Also, we know that

co t0 = ItanOcot 0 = 1-tan 0


We get,

COtQ= 1 34 cot 0 = y4

COt0=43COt© = | .... (4)

We get,

Now by substituting the value of COS0cos ©, tanQtan 0 and COtOcot 0 from equation (2) ,(3) and (4) respectively from the expression below

cos 0 -COS0— 1 tan© 2cotO t a n 0

2 cot 0

We get,

cos 0 -COS0— 1 tan0 2cotO t a n 8 _

-------------- 4 5 -1 3 2 X 4 3 2x!2 cot 0

COS0— 1 tan0 2C0t©C O S 0 -

t a n 0 _

2 cot 01215-2015 83

1215

2015

JS__ 15- -815 83 —

"3

C O S 0 — 7— ^ _ i

Therefore, cos0-itanG2cot0—-— = -15-?-2 cot 0 5

21.) If tan0=247tan@ = y-, find that sin0+cos©sin@ + cos0

Sol.

Given:

tan©=247tan0 = -y .... (1)

To find,

sinO+cosOsin 0 + cos 0

Now we know that tanOtan 0 is defined as follows

tan© — PerpendicularsideoppositetozOBasesideadjacenttozO

tan 0 = Perpendicular side opposite to Z© Base side adjacent to Z© (2)

Now by comparing equation (1) and (2)

We get,

Perpendicular side opposite to Z 0 Z 0 = 24

Base side adjacent to Z 0 Z 0 = 7

Therefore triangle representing Z 0 Z 0 is as shown below

Side AC is unknown and can be found by using Pythagoras theorem

Therefore,

AC2= AB2 + BC2

Now by substituting the value of unknown sides from figure

We get,

AC2= 242 +72

AC = 576 + 49

AC= 625

Now by taking square root on both sides,

We get,

AC = 25

Therefore H hy

Hypotenuse side AC = 25 .... (3)

Now we know sinOsin © is defined as follows

■ ............. . , • r\ Perpendicular side opposite to Z 0S in W - PerpendicularsideoppositetozOHypotenuseSlIl = ---------------- e?—- t-----------------------r Hypotenuse


We get,

sin©=abac s in © =

sin©=2425sin© — .... (4)

Now we know that COS0cos 0 is defined as follows

COS0— BasesideadjacenttozOHypotenuse COS 0 Base side adjacent to Z 0 Hypotenuse

Therefore by substituting the value of sinOsin 0 and COS0COS 0 from equation (4) and (5) respectively, we get

sinO+cosOsin 0 + cos 0 = 2425+ 7 2 5 | | + ^

sinO+cosOsin 0 + cos 0 = 3125

Hence, sinO+cosOsin 0 + cos 0 = 3125

22.) If sinO=absin © = find secO+tanOsec 0 + tan © in terms of a and b.

Sol.

Given:

sin©=absin0 — | .... (1)

To find: secO+tanOsec 0 + tan 0

Now we know, SinOsin 0 is defined as follows

■ „ ................. - / " v Perpendicular side opposite to ZQS in tJ — PerpendicularsideoppositetozOHypotenuseSinfcJ = -------------- ^ — 7--------------------------r Hypotenuse

(2)


We get,

Perpendicular side opposite to Z 0 Z 0 = a

Hypotenuse = b

Therefore triangle representing Z 0 Z 0 is as shown below

Hence side BC is unknown

Now we find BC by applying Pythagoras theorem to right angled AABC A ABC

Therefore,

AC2 = AB2 +BC2

Now by substituting the value of sides AB and AC from figure (a)

We get,

b2 = a2 + BC2

Therefore,

BC2 = b2 - a2

Now by taking square root on both sides

We get,

BC= Vb2- a 2v ^ - a z’

Therefore,

Base side BC = Vb2—a2\/&2- a 2 .... (3)

Now we know COS0cos 0 is defined as follows

COS0— BasesideadjacenttozOHypotenuse COS 0 Base side adjacent to ZO Hypotenuse


We get,

C O S © = BCAC COS 0 =

= Vb2-a2b b

C O S © = BCAC COS 0 = ^

= Vb2-a2b ^ ■■■■ (4)

Now we know, sec© = icosOsec © = —cos 0

Therefore,

sec© = bVb2-a2sec © — . ? .... (5)

Now we know, tan©=sin0cos0tan@ —C O S 0

Now by substituting the values from equation (1) and (3)

We get,

tan©=abvsq?btan© = * tan©=aVb2-a2ta n © = . f

Therefore,

tan©= aVb2-a2ta n 0 — . ° .... (6)

Now we need to find sec0+tan0sec 0 + ta n 0

Now by substituting the values of secQsec 0 and tanQtan 0 from equation (5) and (6) respectively

We get,

sec0+tan0sec 0 + tan © = bVb2-a2 + aVb2-a2 --------- h —t=—V b2 -a 2 \ /b2 -a?

sec0+tan0sec 0 + tan 0 = b+aVb2-a2 h+a .... (7)vV-a2

We get,

Sec0+tan0sec 0 + tan 0 = b+aVb+a-Vb-a /7_ T a/—VO+a-yo-a

Now by substituting the value in above expression

We get,

sec©+tan©sec 0 + tan © = Vb+aWb+aVb+a-Vb-ay o + a -y o -a

Now, Vb+a V ^ -F a present in the numerator as well as denominator of above denominator of above expression gets cancels we get,

sec0+tan0=Vb+aVb::asec0 + tan © =y / b — a

Square root is present in the numerator as well as denominator of above expression

Therefore we can place both numerator and denominator under a common square root sign

Therefore, secO+tanQsVb+aVb^asec© + tan © —y / b — a

23.) If 8tanA=158 tan >1 = 15, find sinA—cosAsin A - cos A

Sol.

Given:

8tanA=158tanA = 15

Therefore,

tanA=i58tanA = ^ .... (1)

To find:

sinA—cosAsin A - cos A

Now we know tan A is defined as follows

tanA— PerpendicularsideoppositetozABasesideadjacenttozA

, a Perpendicular side opposite to A Ata n A = — ^ -----rj— .. w ..------.... (2 )Base side adjacent to /LA v 7


We get

Perpendicular side opposite to zA /A = 15

Base side adjacent to / J K / A = 8

Therefore triangle representing angle A is as shown below

Side AC= is unknown and can be found by using Pythagoras theorem

Therefore,

AC2= AB2 + BC2

Now by substituting the value of known sides from figure (a)

We get,

AC2= 152 + 82

AC2 = 225 +64

AC = 289


We get,

AC = V289\/289

AC = 17

Therefore Hypotenuse side AC=17 .... (3)

Now we know, sin A is defined as follows

_■ * _ . a Perpendicular side opposite to / .AS in A - PerpendicularsideoppositetozAHypotenuseSin A = -------------- tj— ------------------------r Hypotenuse


We get,

sinA=BCACsin^l —

sinA=i5i7sin>l = .... (4)

Now we know, cos A is defined as follows

COSA— BasesideadjacenttozAHypotenuse COS A Base side adjacent to Z A Hypotenuse


We get,

cosA= a ba c c o s A =A C

COSA= 817COS A = .... (5)

Now we find the value of expression sinA— COsAsin A -co s A

Therefore by substituting the value the value of s in A s in A and COsAcos A from equation(4) and (5) respectively , we get,

sinA-cosA=i5i7-8i7sinA-cos A = j|-- jy sinA-cosA=i5-8i7

sin A-cos A = sinA—cosA=7i7sin A-cos A = ^y

Hence, sinA—COsA=7i7sin A - cos A = yy

24.) If tanO= 2021 ta n © = | y , show that 1-sinO-cosOl +sin0+cos0 =37

l-sin 0 -cos 0 _31+sin 0 + cos 0 7

Sol.Given:

tan©= 2021 t a n © -

To show that 1-sin0+cos01 +sin0+cos0 — 37 i1~sin ® +cos ® ^1+sin 0 + cos 0 7

Now we know that

tan© — PerpendicularsideoppositetozOBasesideadjacenttozO

ta n © __ Perpendicular side opposite to ZQBase side adjacent to ZQ

Therefore,

tan©= 2021 ta n 0 =

Side AC be the hypotenuse and can be found by applying Pythagoras theorem

Therefore,

AC2 = AB2 + BC2

AC2= 212 + 202

AC2 = 441 + 400

AC2 = 841


We get,

AC = V 8 4 1 \/8 4 l

AC= 29

Therefore Hypotenuse side AC= 29

Now we know, sinOsin 0 is defined as follows,

■ _ a /''k ................. . . j s-\ Perpendicular side opposite to ZQSinAfcJ- PerpendicularsideoppositetozOHypotenuseSlIl A & — -------------- 77— 7------------------------r Hypotenuse

Therefore from figure and above equation

We get,

sin© = abac sin© = 4 1 sinO= 2029 sin© =

Now we know COS0cos 0 is defined as follows

COS0— BasesideadjacenttozOHypotenuse COS 0 Base side adjacent to Z 0 Hypotenuse

Therefore from figure and above equation

We get,

COS©= ABAC COS 0 = COS©= 2129 COS 0 =A C 29

Now we need to find the value of expression l-sin0+cos0l+sin0+cos0 l-s in 0 + c o s 0 1+sin 0 + c o s 0

Therefore by substituting the value of sinOsin © and COS0cos ©from above equations, we get

1-sin0+cos01 +sin0+cos0 1-sin 0 + c o s 0 _ 1+sin 0 + c o s 0

29-20+2129 7029

29-20+21 297029

Therefore after evaluating we get,

. . _ _ _ l -s in 0 + c o s 0 _ 31-sin0+cos01+sin0+cos0 , , . ------77- 37 ■=1+sin 0 + c o s 0 7

Hence,

1-sin0+cos01 +sin0+cos0 l-s in 0 + c o s 0 1+sin 0 + c o s 0

25.) If COSecA=2cosecA = 2 , find 1tanA+sinA1+ c o s A ^ ^ - + 1 '^ £ A

Sol.

Given:

cosecA=2 cosecA — 2

To find 1 tanA + sinA1 +cosA -r~~r +tan. x\. x | cos

U ypotenuseNow cosec A = HypotenuseOppositeside —------ -— rr-Uppositesiae

Here BC is the adjacent side,

By applying Pythagoras theorem,

AC2= AB2 + BC2

4 = 1 + BC2

BC2= 3

BC = V3\/3

Now we know that

s in A = ico secA s in A = —cosecA

sinA=i2 sinA = .... (1)

tanA=ABBCtan A — 45

tanA=iV3tanA = -j= .... (2)

c o s A = b ca cc o s A = AC

cosA= V32 COS A = ^ .... (3)

Substitute all the values of sinAsin A , COsAcos A and tanAtan A from the equations(1), (2) and (3) respectively

We get.

i1tanA + sinA1+cosA — ^ - r + sin^- r = 11V3 + 12I+V32 —} -----1------

tan A 1+cos A , v3V3 1 + ~

-V3+12W3x/ 3 + ^

= 2(2+V3)2+V32(2+ V 3)

3 2+V3

= 2

Hence,

tan A 1+cos A

26.) If Z A Z A and Z B Z S are acute angles such that cos A =cos B , then show thatzA Z A = Z B LB

Sol.

Given:

Z A Z A and zB Z i? are acute angles

cos A = cos B such that Z A Z A = zB Z i?

Let us consider right angled triangle ACB

A

c B

Now since cos A = cos B

Therefore

ACAB = BCAB ^ =AC _ BCAB AB

Now observe that denominator of above equality is same that is AB

Hence ACAB = BCAB only when AC=BCAB AB J

Therefore AC=BC

We know that when two sides of triangle are equal, then opposite of the sides are also

Equal.

Therefore

We can say that

Angle opposite to side AC = angle opposite to side BC

Therefore,

z B Z 5 = z A Z A

Hence, Z A /.A = Z B /.B

27.) In a AABCAABC , right angled triangle at A, if tan C = V3-\/3 , find the value of sin B cos C + cos B sin C.

Sol.

Given:

A A B C A A B C

To find : sin B cos C + cos B sin C

The given a A A B C A A B C is as shown in figure

Side BC is unknown and can be found using Pythagoras theorem,

Therefore,

BC2= AB2 +AC2

BC2=^l32y/32 + 12 BC2 = 3+1

BC2 = 4


We get,

BC = V 4 i/4

BC= 2

Therefore Hypotenuse side BC= 2 .... (1)

Now, sin B = PerpendicularsideoppositetozBHypotenuse Perpendicular side opposite to ZB Hypotenuse

Therefore,

sinB=ACBCsin£ = ^

Now by substituting the values from equation (1) and figure

We get,

sin B = 12 \ ....(2)

Now, COS B= basesideadjacenttozBHypotenuse base side adjacent to ZB Hypotenuse

Therefore,

cos B = ABBC BC

Now substituting the value from equation

cos B= V 3 2 ^ .... (3)

Similarly

sin C = V 3 2 ^ ....(4)

Now by definition,

tanC=sinCcosCtanC =

So by evaluating

COsC=12cosC = \ .... (5)

Now, by substituting the value of sinB, cosB.sin C and cosC from equation (2) ,(3) ,(4) and(5) respectively in sinB cosC + cosB sin C

sinB cosC + cosB sin C= 12X12 + V32XV32 ^ x \ + ^ x

= M + 3 4 j + |

= 1

Hence,

sinB cosC + cosB sin C = 1

28.) State whether the following are true or false. Justify your answer.

(i) The value od tan A is always less than 1.

(ii) sec A = 125 ^ for some value of Z A Z A

(iii) cos A is the abbreviation used for the cosecant of Z A /LA.

(iv) s in 0=43s in 0 = for some angle 0 0 . Sol.

Sol.

(i) tan A < < 1

Value of tan A at 45° i.e... tan 45=1

As value os A increases to 90°

Tan A becomes infinite

So given statement is false.

(ii) sec A = 125 -y for some value of angle if

M-l

sec A = 2.4

sec A > 1

M - II

For sec A = 125 ^ we get adjacent side = 13

Subtending 9i at B.

So, given statement is true.

(iii) Cos A is the abbreviation used for cosecant of angle A.

The given statement is false.

As such cos A is the abbreviation used for cos of angle A , not as cosecant of angle A.

(iv) Cot A is the product of cot A and A

Given statement is false

cot A is a co-tangent of angle A and co-tangent of angle A = adjacentsideOpositeside

adjacent side * Sol.

Oposite side

(v) s in© =43sin© — -f for some angle 0 0 .o

Given statement is false

Since value of sinOsin 0 is less than(or) equal to one.

Here value of sinOsin 0 exceeds one,

So given statement is false.

So given statements is true.

29.) If SinO= 1213sin0 = -j| find sin20-cos202sin0cos0Xltan20 sin2 0 —cos2 0 12 sin 0 cos 0 tan2 0

Sol.

Given: S in © = l2 l3 s in 0 = -j|

To Find: sin20-cos202sin0cos0 * 1tan20sin2 0 —cos2 0 12 sin 0 cos 0 tan2 0

As shown in figure

Here BC is the adjacent side,

By applying Pythagoras theorem,

AC2=AB2+BC2

169 = 144 +BC2

BC2= 169-144

BC2= 25

BC = 5

Now we know that,

COS0— basesideadjacenttozOHypotenuse C O s Q

C O S 0 = cos0= 513 COS 0 —AC 13

base side adjacent to ZQ Hypotenuse COS0= BCAC

We also know that,

tan0=sinOcosOtan0 = :

Therefore, substituting the value of sin0sin 0 and COS0cos © from above equations

We get,

tan 0 = 1 25 tan© = ^

Now substitute all the values of sinOsin © , COsOcos 0 and tanOtan © from above

equations in sin2Q-cos2Q2sinQcosQ * 1tan2Q s*n. ®~cos ® x — \ —2 sin 0 cos 0 tan2 0

We get,

sin* 20 -cos202sin0cos0 x 1tan20 sin2 0 -c o s 2 0 2 sin 0 cos 0

Xtan2 0

2 2 2 (1213) - ( 513 ) 2 x (1213)x (5 1 3 )X 1 ( 125 )

a t r - g y :: ,KsM*) (f)!Therefore by further simplifying we get,

sin20 -cos202sin0cos0 x 1tan20 s™2. 9 cos2 ® x — = 119169 x 169120 x 251442 sin 0 cos 0 tan2 ©

119 ^ 169 25169 120 144

Therefore,

sin20 -cos202sin0cos0 x 1tan20 s*n . ® cos ®2 sin 0 cos 0X

1tan2 ©

= 5953456 5953456

Hence,

sin20-cos202sin0cos0 x 1tan20 sin2 0 —cos2 0 2 sin 0 cos 0

X 1tan2 0

= 5953456 5953456

30.) If COS©— 513 COS © = Tjj, find the value of sin20-cos202sinOcosO * 1tan20

sin2 0 —cos2 0 12 sin 0 cos 0 tan2 0

Sol.

Given: If COS0=513COS© = 7^

To find:

The value of expression sin20 -cos202sin0cos0 x ltan20 s*n. cos x2 sin 0 cos 01

tan2 0

Now we know that

COS ©COS© = basesideadjacenttoz0Hypotenusebase side adjacent to Z©

Hypotenuse (2)

Now when we compare equation (1) and (2)

We get,

Base side adjacent to Z 0 Z © = 5

Hypotenuse = 13

Therefore, Triangle representing Z0Z@ is as shown below

Perpendicular side AB is unknown and it can be found by using Pythagoras theorem


We get,

AC2= AB2 + BC2

Therefore by substituting the values ogf known sides,

AB2 = 132 - 5 2

AB2= 169 -25

AB2 = 144

AB = 12 .... (3)

Now we know from figure and equation,

sin©=i2i3sin@ — .... (4)

Now we know that,

tan0=sin0cosoton© =cos 0

tan © = i25tan® — ^ .... (5)

Now w substitute all the values from equation (1), (4) and (5) in the expression below,

sin20 -cos202sin0cos0 x 1tan20 S*D. ®~cos ® x — \—2 sin 0 cos 0 tan 0

Therefore

We get,

( f y - u r :: ,

>x(s)-(i) ( f )2T h e r e fo r e b y fu r th e r s im p lify in g w e g e t,

sin20 -c o s 20 2 s in 0 c o s 0 x ita n 20 ^ 9 ~ cos2 ® x — = 1 1 9 1 6 9 X 1 6 9 1 2 0 X 2 5 1 4 42 sin 0 cos 0 tan2 ©

119 ^ 169 25169 120 144

T h e r e fo r e ,

sin20 -cos202sinOcosO x itan20 ^ 9 ~cos2 ® x — = 5953456 tS2 sin 0 cos 0 tan2 © 3456

H e n c e ,

. sin2 0 —cos2 0 w 1 _______ ___ 595sin 0-cos202sin0cos0Xitan20 - r — ----- :r— X — 5— = 59534562 sin 0 cos 0 tan2 0 3456

31.) If sec A = 178 , verify that 3-4sin2A4cos2A-3 = 3-tan2A1-3tan2A

3 -4 sin2 A __ 3 -tan2 A4 cos2 A - 3 1 -3 tan2 A

Sol.

G iv e n : s e c A = 1 7 8 ^

T o v e r ify : 3 -4 s in 2A4cos2A -3 = 3 -ta n 2A 1 -3 ta n 2A 3 4 s„in A = 3 tan ^3 4 cos2 A - 3 1 -3 tan2 A

N o w w e k n o w th a t C O S A = is e c A c o s A = —sec A

N o w , b y s u b s titu t in g th e v a lu e o f s e c A

W e g e t ,

C O S A = 817C O S A — JY

N o w w e a ls o k n o w th a t,

sin2A+cos2A= 1 sin2 A + cos2 A — 1

Therefore

• 2 2 a 1 2 2 2sin2Q-cos2Q2sinQcosQ x 1tan20 S*D. _ cos " X — = (1213) - ( 513) 2x(i2i3)x(5i3)x 1(125)2 sin 0 cos 0 tan2 0

sin2A=1-cos2Asin2 A = 1-cos2 A

= <817)2( w )2

= 225289 225289

Now by taking square root on both sides,

We get,

sinA=i5i7sin>l —

We also know tha t, tanA= sinAcosA ta n A = sin A cos A

Now by substituting the value of all the terms ,

We get,

tanA=i58tan>l = ^

Now from the expression of above equation which we want to prove:

L.H.S = 3-4sin2A4cos2A-3 4 sin \4 cos2 A - 3

Now by substituting the value of cos A ad sin A from equation (3) and (4)

We get,

3 \ 225L.H.S = 3 -42252894—64289-3 ------

4-^r-3

= 867-900256-867

= 33611 m

From expression

R.H.S = frac3-tan2A1-3tan2A/rac3-tan2 A l - 3 t a n 2 A

Now by substituting the value of tan A from above equation

We get,

R.H.S= 3 —(158) 1 -3 (1 5 8 ) ------------------2

= -3364-6116464

-61164

= 33611 m

Therefore,

We can see that,

3—4sin2A4cos2A-3 = 3-tan2A1-3tan2A 3-4 s0in2 A — 3~tan2,A4 cos2 A - 3 1 -3 tan2 A

32.) If sinO=34sin© = j , prove that Vcosec20-cot20sec2-1 = \7 3

cosec2 Q -cot2 9 -\/7sec2- l 3

Sol.

Given: s in © = 3 4 s in 0 = | .... (1)

To prove:

Vcosec2O^ot20sec=M = V/3 [ c P^c2Q-cot2 9 = ^ (2)V sec2- l 3

By definition,

sin A = PerpendicularsideoppositetozAHypotenuse Perpendicular side opposite to / .A Hypotenuse


We get,


Hypotenuse = 4

Side BC is unknown.

So we find BC by applying Pythagoras theorem to right angled AABCAABC

Hence,

AC2 = AB2 +BC2

Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)

Therefore,

42= 32 +BC2

BC2 = 1 6 - 9

BC2= 7

BC = yfry/7

Hence, Base side BC =V7\/7 .... (3)

Now cos A = bcac4 ^

V74^ ....(4)

N o w , c o s e c A = i s i n A cosec A — - 4 - rsin A

Therefore, from fig and equation (1)

COSecA— HypotenusePerpendicular COSec.A = Hypotenuse

Perpendicular

cosecA= 43cosecA = | .... (5)

Now, similarly

secA=4\7sec A. = ..(6)

Further we also know that

COtA— cosAsinA cot A = ^ 4sm A

Therefore by substituting th values from equation (1) and (4),

We get,

COtA=V73COt A = .... (7)

Now by substituting the value of cosec A, sec A and cot A from the equations (5), (6), and(7) in the L.H.S of expression (2)

V c o s e c 2 0 ^ o t2 0 s e c ^ 1 / ^ c 20 -c o t2 0 = m n ^ lV sec2- l

n n n N (43) -(V73) (4V7) -1 ( D 2- ( ^ ) ;N B ) 2- '

= 1 6 9 -7 9 1 6 7 -1

169 9

Hence it is proved that,

Vcosec2€)-cot20sec2-1 = -.73 9 = ^V sec2- l o

33.) If SecA=178secA = ^ , verify that 3-4sin2A4cos2A-3 = 3-tan2A1-3tan2AO3 -4 sin2 A 3—tan2 A4 cos2 A - 3 1 -3 tan2 A

Sol.

Given: secA=i78secA = ^ .... (1)

To verify:

3-4sin2A4cos2A-3 = 3-tan2A1-3tan2A 3 4 sm A = 3 t a n .... (2)4 cos2 A -3 1 -3 tan2 A v ’

Now we know that sec A = icosA — —rcosA

Therefore COSA=lsecAcos A =secA

We get,

COSA=817COS A = -pf .... (3)

Similarly we can also get,

sin A= s in A = l5 l7 s inA = ■■■■(4)

An also we know that tanA= sinAcosA tan A — - ^ 4

tanA=i58tan>l = ^ .... (5)

Now from the expression of equation (2)

L.H.S: Missing close brace Missing close brace

Now by substituting the value of cos A and sin A from equation (3) and (4)

We get,

2 2 3 -4L.H.S = 3-4(1517) 4(817) -3 (# ) '

-3

- 867-900289256-867289

867-900289

256-867289

= 33611 J r . . . . (6 )611

R.H.S = 3-tan2A1-3tan2A 3—tan2 A 1-3 tan2 A

Now by substituting the value of tan A from equation (5)

We get,

R.H.S=3-(1518) 1-3(158) ,

-3364

-3364 -61164 — ^

64

= 33611 J r . . . . (7 )611


We get,

3-4sin2A4cos2A-3 = 3-tan2A1-3tan2A s„in An4 cos2 A - 3

3—tan2 A 1-3 tan2 A

sec Q cosecQ 1sec Q +cosecQ y/7

Sol.

Given: COt©=34cot@ — -74

34.) If COt0= 34cot 0 = -|, prove that sec0-cosec0sec0+cosec0 = lV7

Prove that: sec0-cosec0sec0+cosec0 — 1V7 sec ® cogec®sec 0 + c o se c 0

1_^7

Now we know that

sec0-cosec0sec0+cosec0 — sec 0 -c o se c 0 sec 0 + c o se c 0

1

Here AC is the hypotenuse and we can find that by applying Pythagoras theorem

AC2= AB2 +BC2

AC2 = 16 +9

AC2= 25

AC = 5

Similarly

sec0=ACBCsec0 — ^ sec0=53sec0 — | cosec=ACAB cosec — 4§ cosec=54AB

cosec = j

Now on substituting the values in equations we get,

sec0-cosec0sec0+cosec0 — 1V7 sec®~^osec®sec 0 + c o se c 0 y 7

Therefore,

sec0-cosec0sec0+cosec0 = 1V7 sec ® cosec®sec 0 + c o se c 0

1

Sol.

Given: 3cosO—4sin©=2cos©+sin©3 cos 0 -4 sin 0 = 2 cos 0 + sin 0

To find: tanOtan 0

We can write this as:

3cos0—4sin0=2cos0+sin03 cos 0 - 4 sin 0 = 2 cos 0 + sin 0 cos©=5sin©cos 0 = 5 sin 0

Dividing both the sides by COS0cos 0 ,

We get,

cosOcos©=5 s i n © c o s 0 = 5sin® 1=5tan©l = 5tan0 tanO=1tan© = 1cos 0 cos 0

Hence,

tanO=1tan 0 = 1

35.) If 3cosO—4sinO=2cosO+sin03 cos 0 - 4 sin 0 = 2 cos © + sin 0 .find tanOtan©

36.) If aAZA and ZP Z P are acute angles such that tan A = tan P, then show

zA = zP Z .A = Z P

Sol.

Given: A and P are acute angles tan A =tan P

Prove that: zA = zP Z A = Z P

Let us consider right angled triangle ACP

W e know t a n G —oppositesideadjacentsidetan© =

P C

tan A = pcac^77AC

tan P =a c pc^

tan A =tan P

PCAC = ACPC PC AC AC PC

PC =AC [v.-Angle opposite to equal sides are equal]

Z Z A A = Z P


Evaluate each of the following:

Q 1 . sin 45°45° sin 30°30° + cos 45°45° cos 30°30°

Solution:

Sin 45°45°sin 30°30° + cos 45°45° cos 30°30°

[1]

We know that by trigonometric ratios we have ,

sin45°=iV2sm45° =v 2

Sin30°=i2sm30°

cos45°=i\2 cos45° = A=V2 cos30°=V32cos30

Substituting the values in equation 1 , we get

W5-12+W5-V32-L . 1 + ^ . 4

Q 2 . sin 60°60° cos 30°30° + cos 60°60° sin 30°30°

Solution:

sin 60°60° cos 30°30° + cos 60°60° sin 30°30°[1]

By trigonometric ratios we have ,

sin60°=V32sm60° = ^ Sin30°=i2sm30° = \

cos30°=V32cos30° = ^ cos60°=12cos60° = \


= V32-V32 + 1 2 - 1 2 ^ ' ^ + 2 ‘ 2

= 34+ 14 4 + T = 44 4 = 14 4 4

Q 3 . cos 60°60° cos 45°45° - sin 60°60° sin 45°45°

Solution:

cos 60°60° cos 45°45° - sin 60°60° sin 45°45°[1]


cos60°=12cos60° = \ cos45°=iV2cos45°

sin60°=V32sm60° — ^ sin45°=iV2sm45° — -±=£ \/2


12-1V2-V32-1V2-| • V3 _ J _ 2 ’

= 1-V32V2 1—\/32 /2

Q.4: sin230o+sin245o+sin260o+sin290osin230o + sm245° + sin260° + sin290c

Solution:

sin230o+sin245°+sin260o+sin290°sm230° + sin245° + sin260° + sin2 90° [1]


Sin30°=i2sm30° = \ sin45°=iV2sm45° = V2

sin60°=V32sm60° = ^ sin90°sm90° = 1


= [12]2+ [W 2] +[V32] + 1 [ | ] 2 +n 2

L V2 J + V3 + 1

= 14 + 12 + 34 + 1 ^ + 1 + 1 + 1

= 52 —“ 2

Q 5. cos230°+cos245°+cos2600+cos2900cos2 30° + cos245° + cos2 60° + cos2 90°

Solution:

cos2300+cos2450+cos2600+cos290°cos2 30° + cos245° + cos2 60° + cos2 90° [1]


cos30°=V32cos30° = ^ cos45°= i V2cos45° = - j -

cos60°=12cos60° = \ cos90°cos90° = 0


[V32] +[1V2] + [ 1 2 ] +0 V?2

1 2+ 1

V* + [ l ] 2 + o

34 + 12 + 14| + \ + \

321

Q 6. tan230°+tan245o+tan260°£an230o + t a n 2 45° + t a n 2 60c

Solution:

tan230o+tan245°+tan260°tan230o + t a n 2 45° + t a n 2 60°[1]


tan30°=iV3tan30° — -7 = tan60°=V3£an60° — \/3

tan45°=1tan45° = 1


2 - 2[iV3f+[V3f+1 “ + [V 3 ]2 + 1

= i3+3+1-| + 3 + l

Q 7 .2sin230o-3cos245°+tan260o2sm230° - 3cos245° + tan260°

Solution:

2sin230°-3cos2450+tan26002sm230° - 3cos245° + tan260°


Sin30°=i2sm30° — \ cos45°= i V2cos45° — - 7=

tan60°=V3fcm60° = \/3


= 2(i2)2-3(W i)2+(V3)22 (D 2 - 3 ( ^ ) 2 + (V3)2

= 2 ( i4 ) -3 ( i2 )+ 3 2 ( |) -3 ( i )+ 3

= 2

Q8:sin230°cos245°+4tan230°+ i2sin290°-2cos290°+ i24cos20°

sm230°cos245° + 4tan230° + |s m 290° — 2cos290° + ^cos20°

Solution:

sin230°cos245°+4tan230°+ i2sin290°-2cos290°+ i24cos20°

sm230°cos245° + 4tan230° + -|sm290o — 2cos290° + ^cos2 0°[1]


Sin30°=i2sm30° = | cos45°=iV2cos45° = tan30°=iV3tan30°

sin90°sm90° = 1

cos90°cos90° = 0

cos0°cos0° = 1


2[0]2 + £ [ l ]2

= 18 + 43 + 1 2 + 1 2 4 i + | + i + ±

= 4824 | | = 2

Q 9 .4(sin460°+cos4300)-3(tan2600-tan2450)+5cos245°4 (sm 460° + cos430°) — 3 ( t a n 2 6 0° — t a n 2 45°) + 5cos245°

Solution:

4(sin460°+cos4300)-3(tan2600-tan2450)+5cos245°4 (sin460° + cos430°) — 3 { t a n 2 60° — t a n 2 45°) + 5cos245°[1]


sin60°=V32sm60° = ^ cos45°=iV2cos45° = -4-2 ^2

tan60°=V3fan60° = a/3 cos30°=V32cos30° = ^


4([V32] +[V32] )-3 (3 )-1 2+5[iV2] 4 r + 5 i_V2

= 4-1816-6+524 • -ff — 6 + I

= 14-6+52 | - 6 + |

= 1 4 2 -6 ^ - 6 = 7 - 6 = 1

Q 10. (cosec245°sec2300)(sin2300+4cot2450-sec260°)(cosec2 45° sec2 30°) (s in2 30° + 4cot245° — sec2 60°)

Solution:

(cosec245°sec2300)(sin2300+4cot2450-se c260°)(cosec2 45° sec2 30°) (s in2 30° + 4cot245° — sec2 60°)

[1]


cosec45°=V2cosec45° = y / 2 sec30°=2V3sec30° =

Sin30°=i2sin30° = \ cot45°cot45° = 1

sec60°sec60° = 2


- 2([V2] .[*& ] ) ( [ i2f +4(1)(2)2) [^2] 2

v^J+ 4(1) (2)2)

= 3-43-143 • | • 14

Q11. cosec3300cos60°tan3450sin2900sec2450cot300cosec3 30° cos60° tan3 45° sin2 90° sec2 45° cot30°

Solution:

Given,

= cosec330°cos600tan3450sin2900sec2450cot30° cosec3 30° cos60° tan3 45° sin2 90° sec2 45° cot30°

= 23(i2 )(13)(12)(V22)(V3)23( i ) ( l 3) ( l 2) ( ^ 2)(V 3 )

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x (g

i) x

(|)

x E

(z)(

e/v

)x(^

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Ox(

Ei.)

x(sO

xete

) =

(cos0°+sin450+sin300)(sin900-cos450+cos60°)(1 + 1V2 + 1V2 )(1—1V2 + 1V2 )

( 32 + 1V2 )( 32 — 1V2 ) (( 32 )2—(1 V2 )2) 94 — 12 74

(cosO° + sm45° + sm30°)(sm90° — cos45° + cos60°)

f 1 + 75 + + 75^

( ( 3\2 _ Z J_^2\ 9 _ 1 7 U 2 ^ ' ^/2 ' ' 4 2 4

Q14. sin30°-sin90°+2cos0°tan30°tan60° sin30°—sin90°+2cos0°tan30°tan60°

Solution:

Given,

sin30o—sin90°+2cos0°tan30°tan60°

_ Osin30°-sin900+2cos00tan300tan60° 12-I+ 21V3W3 32 ^

Q15. 4cot230° + 1sin260°—COS245°— } ------h ~ r 4 ----------COS245°cot2 30° sin2 60°

Solution:

Given,

4cot230° + 1 sin260° -C O S 24 5 ° = 4(V3)2 + 1 (V32)2—( 1<2 ) 2= 43 + 43 - 12 = 16-36 = 136

cot2 30c +

(V3)2+

sin2 60° 1

( f ) !4 1 4 _ 1 3 ' 3 216-3

- C O S 2 45°

136

Q16.4(sin4300+cos260°)-3(cos2450-sin2900)-sin26004(sm430° + cos26 0 ° ) — 3(cos245° — sin2 90°) — sin2 60°

Solution:

Given,

4(sin430°+cos2600)-3(cos2450-sin2900)-sin260°=4(( 12 )4+( 12 )2)-3(( 1V2 )2-1 )-

( V32 )2=4( 116 + 1 4 )+ 32 — 34 = 84 =2

4(sm430° + cos26 0 ° ) — 3(cos245° — sin2 90°) — sin2Q0° = 4 ( ( i)4 + (D 2) - 3 ( ( ^ ) 2 - l ) - ( # ) 2

= 4 ( ^ + 1) + ! - !_ 8 _ o

Q17. tan260°+4cos245°+3sec230°+5cos290°cosec30°+sec60°-cot230'

tan2 60°+4cos245°+3sec2 30°+5 cos2 90° cosec30°+sec60°—cot2 30°

Solution:

Given,

tan260°+4cos2450+3sec2300+5cos2900cosec300+sec60°-cot2300 = (V3)2+4(iV2)2+3(2V3)2+5(0)2+2-

tan260°+4cos245°+3sec230°+5cos290° cosec30°+sec60°—coi2 30°

_ (a )2+ 4 (^ )2+ 3 (^ )2+5(0)

2 + 2 -(V 3 )2

= 3 + 2 + 4 (V3)2=3+2+4=9= 9

Q18. sin30°sin45° + tan45°sec60°—sin60°cot45°—cos30°sin90°smA5

I tan45° sin60°”l~ sec60° coM5°

Solution:

cos30°sm90°

Given,

sin30°sin45o + tan45osec60°—sin60°cot45°—cos30°sin90° — 121V2 + 1 2 —V32I — V32I — V22 + 12—V32

sin30° 1 sinAh0

tan45° sm 60° cos30'sec60° cot 45° sin90

y3

J _ 1 2 1 1V2

V32 = V2+1-2V32 2

, 1_ ^3_ v f2 ' 2 2 2

^+1-2^3

1 i2 ' 2

Q 19. tan45°cosec30° + sec60°cot45° + ssin90°2cos0 tan45° + sec60'cot45'

+ ssin90'2 cos0 °cosec30

Solution:

Given,

tan45°cosec30° + sec60°cot45° + ssin90°2cos0° = 12 + 21-5(1 )2(1) = 5 2 -5 2 =0tan45° , sec60° , ssin90°

cosec30° cot45° 2cos0°

_ 1 , 2 5(1)2 1 2(1)5 _ 5 2 2

= 0

Q20.2sin3x=V32sm3cc = y/%

Solution:

Given,

2sin3x — y/3

= > sin3x — y

— > sin3x = sin60° = > 3 x = 60°

2sin3x=V3=>sin3x=V32=>sin3x=sin60°=>3x=60°=>x=20o=> x = 20°

Q 21)2s in x2= 1 ,x= ? 2sm | = 1, x =?

Solution:

sinx2 = i2 sm | = \ sinx2=sin30°sm§ = sin30° x2=30°f = 30°

x = 60°

Q22) V3sinx=cosxv^3sma: = cosx

Solution:

V3tanx=1\/3 tanx = 1 tanx=iV3tanx = .’.tanx=tan45°.\ tanx = £an45°

x = 45°

Q23) Tan x = sin 45° cos 45° + sin 30°

Solution:

T anx= 1V2.1V2+i2[vsin45°= iV2COs45°= iV2sin30°= 12]

Tanx — [y sm45° — cos45° — sm30° = -|] T a n x = i2 + i 2

Tanx = ^

Tan x = 1

Tan x = 45°

x = 45°

Q24) V3Tan2x=cos60°+sin450cos45°\/3 Tan2x = cos60° + sm45°cos45°

Solution:

V3T an2x= 1 2 + 1V2 . 1V2 [vcos60°= i2sin45°=cos45°= 1 V2 ]

y/3 Tan2x — I + [v cos60° — sin45° — cos45° — ]

V3T an2x=iV3=>tan2x=tan30°v/3 Tan2x = -j= ^ tan2x = tan30°

2x = 30°

x = 15°

Q25) cos2x=cos600cos300+sin60°sin300cos2x — cos60° cos30° + sm60°sm30°

Solution:

cos2x= 1 2 . V3 2 +V3 2 . 12 [vcos60°=sin30°= I2sin60°=cos30°= V3 2 ]

cos2x — I . ^ + "T"- \ [’•’ cos60° — sm30° — -|sm60° — cos30° —cos2x=2.V34cos2a: — 2 .^ cos2x=l32cos2x — ^ cos2x=cos30°cos2a; — cos30°

2x=30°2a: = 30° x=15°:c = 15°

Q26) I f 6 = 30°, verifylf0=3O°,verify(i)T an20=2Tanei-tan2e(i)Tan20 = -

Solution:

T an20=2Tanei-tan2e ..... (\)Tan20 = .......(i)

Substitute 0=30°0 — 30° in equation (i)

LHS = Tan 60° = V3V3

_ 2__LR H S = 2T an30°1 + (T an30°)2 = 2 -1 Vi 1 - ( 1V2)2 = V 3 2Tan30° = -------- = ^ / 3

l+ (T a n 3 0°)2 l - ( - L ) 2V 2

Therefore, LHS = RHS

(ii) sin0=2tanei-tan 2%sin0 = 2tane0n1 —tan*0

Substitute 0=3O°0 = 30°

S in 6 0 0=2tan30°(1-tan30°)2s in 6 0 0 = 2ton30°n .(1—tan 30°)2

— 2 —1-=>V32 = 2.iVi1+(iVi)2 ^ — ------

2 !+ ( ^ ) 2

= > V 3 _ 2 32 y/3 m 4

— .V32 = 2V3.34 => V32 = V32 ^ ~2~ — ~2~

Therefore, LHS = RHS.

(iii) COS20=1-tan2ei+tan2ecos20 1— tan2 6

1 +tan 2 0


LHS = cosec0cosec0 RHS = 1-tan201+tan20 1—tan2 6

1+tan 2 6

= cos 2(30°) = 1 -ta n 230°1 +tan230° =

Cos 60°= 12 — 1 —(1V2)21 +(1V2)2 — 22 12 — 12 =1+( ^ ) 2

1—ton230°

l + i a n 230°

22_ _ X1 — 22


(iv) c o s 36=4c o s36 -3 c o s6cos30 = 4 c o s 3 6 — 3 c o s 0

Solution:

LHS = cos30cos30


= cos 3 (30°) = cos 90°

= 0

RHS = 4cos30-3cos04cos30 - 3cos6

= 4cos330°-3cos30°4cos330° - 3cos30°

= 4 (V 32 )3- 3 . V 3 2 4 ( ^ ) 3 - 3 . ^

= 3 -V 3 2 -3 .V 3 2 3 .4 -3 .4

= 0

Therefore, LHS = RHS.

Q27) If A = B = 60°. Verify (i) Cos (A - B) = Cos A Cos B + Sin A Sin B

Solution:

Cos (A - B) = Cos A Cos B + Sin A Sin B....... (i)

Substitute A and B in (i)

=>cos (60° - 60°) = cos 60° cos 60° + sin 60° sin 60°

=>COS 0° = (12)2+(V32)2( i ) 2 + ( ^ ) 2

=>1= 14 + 3 4 | + |

= >1 = 1Therefore, LHS = RHS

(ii) Substitute A and B in (i)

=>sin (60° - 60°) = sin 60° cos 60° - cos 60° sin 60°

=> sin 0° = 0

=>0 = 0


(iii) T a n (A -B )= T an A -T an B 1 + T an A T an B T an (4 - B) = anATanB

A = 60°, B = 60° we get,

T a n ( 6 0 0 —6 0 ° ) = T a n 6 0 °-T a n 6 0 °1 + T a n 6 0 °T a n 6 0 °T £ m ( 6 0 ° - 6 0 ° ) = rffln6°°

Tan 0° = 0

0 = 0


Q28 ) If A = 30°, B = 60° verify:

(i) Sin (A + B) = Sin A Cos B + Cos A Sin B

Solution:

A = 30°, B = 60° we get

Sin (30° + 60°) = Sin 30° Cos 60° + Cos 30° Sin 60°

Sin (90°)= 12.12 + V32.V32-|.-| +

Sin (90°) = 1 => 1 = 1


Tan60°) Tan60°

(ii) Cos (A + B) = Cos A Cos B - Sin A Sin B

A = 30°, B = 60° we get

Cos (30° + 60°) = Cos 30° Cos 60° - Sin 30° Sin 60°

Cos (90°)= 12 .V 32 -V 32 .12^ \

0 = 0


Q29. If sin(A+B) = 1 and cos(A-B) = 1, 0°<A+B<90°0° < A + B < 90°, A>B find A and B.

Sol:

Given,

sin(A+B) = 1 this can be written as sin(A+B) = sin(90°)sm(90°)

cos(A-B) = 1 this can be written as cos(A-B) = COS(0°)cos(0°)

=> A + B = 90°90°

A - B = 0°0°

2A = 90°90°

. Qf1°A = 90°2^-

A = 45°45°

Substitute A value in A - B = 0°0°

45°45° - B = 0°0°

B =45°45°

Hence, the value of A = 45°45° and B =45°45°

Q30. If tan(A-B) = 1V3 4 * and tan(A+B) = V3a/ 3, 0°<A+B<90°0° < A + B < 90°,

A>B find A and B

Solution:

Given,

tan(A-B) = 1V3-^

A - B =tan 1 ( 1V3 )tan~1(-^)

A - B = 30°30° ------1

tan(A+B) =

A + B = tan_1V3tan_l\/3

A + B = 60°60° -------2

Solve equations 1 and 2

A + B = 30°30°

A - B = 60°60°

2A = 90°90°

A = 90°2 ^

A = 45°45°

Substitute the value of A in equation 1

45°45° + B = 30°30°

B = 30°30° - 45°45°

B = 15°15°

The value of A = 45°45° and B = 15°15°

Q31. If sin(A-B) = 12 \ and cos(A+B) = 12-|, 0°<A+B<90°0° < A + B < 90°, A<B

find A and B.

Solution:

Given,

sin(A-B) = 12-|

A - B = sin_1( i2)sm_ l(-|)

A - B = 30°30° -------1

cos(A+B) = 12-|

A + B = COS-1 (12 )cos_1( | )

A + B = 60°60° -------2


A + B = 60°60°

A - B = 30°30°

2A = 90°90°

A = 90°2 ^

A = 45°45°

Substitute the value of A in equation 2

45°45° + B = 60°60°

B = 60°60° - 45°45°

B = 15°15°

The value of A = 45°45° and B = 15° 15°

Q32. In a A A ABC right angled triangle at B, ZA=Z.CZA ZC.Find the values of:

1. sinAcosC + cosAsinC

Solution:

since, it is given as zA=zC/.A — Z.C

the value of A and C is 45°45°, the value of angle B is 90°90°

because the sum of angles of triangle is 180°180°

=> sin(45045°)cos(45°450) + cos(45045°)sin(45045°)

=> x - i ) + ( i ^ x ^ X - i x -3 j)

=> 12-| + 12 i

= > 1

The value of sinAcosC + cosAsinC is 1

2. sinAsinB + cosAcosB

Solution:

since, it is given as zA=zCZ_A = Z.C

the value of A and C is 45°45°, the value of angle B is 90°90°

because the sum of angles of triangle is 180°180°

=> sin(45°45o)sin(90°90o) + cos(45°45o)sin(90°90o)

=> WS-^(O)

=> 1 ^ ^ + o

= > 'f2T2

The value of sinAsinB + cosAcosB is 1V2 -4=y/2

- /3Q33. Find the acute angle A and B, if sin(A+2B) = V32 and cos(A+4B) = 0, A>B.

Solution:

Given,

sin(A+2B) = V 3 2 ^

A + 2B = Sin-1 V32 sin~1 -

A + 2B = 60°60° ------- 1

Cos(A+4B) = 0

A + 4B = s i r r 1(90 )sm _ l (90)

A + 4B = 90°90° ------- 2


A + 2B = 60°60°

A + 4B = 90°90°

(-) (-) (-)

-2B = -30°30°

2B = 30°30°

B = 30°2

B = 15°15°

Substitute B value in eq 2

A + 4B = 90°90°

A + 4(15°15°) = 90°90°

A + 60°60° = 90°90°

A = 90°90° - 60°60°

A = 30°30°

The value of A = 30°30° and B = 15°15°

Q 34. In AP QRAPQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine Z Z P

and Z Z R.

Solution:

Given,

In APQRAPQR, right angled at Q, PQ = 3 cm and PR = 6 cm


PR2=PQ2+QR2=>62=32+QR2=>QR2=36-9=>QR=V27=>QR=3V3PR2 = PQ2 + QR2 = > 62 = 32 + QR2 = > QR2 = 36 -9

= > QR = y/27 —> QR — 3 /3

sin R = 36 = I2=sin30°| — — sm30°

zR=30°Zi? = 30°

As we know, Sum of angles in a triangle = 180

zP +zQ+zR= 180°=>zP +90°+30°=180°=>zP=180°-120°=>zP =60°Z P + Z Q + Z i* = 180°= > Z P + 9 0 ° + 30° = 180°= > Z P = 180°-120°= > Z P = 60°

Therefore, zR=30°ZP = 30°

And, z P = 6 0 °Z P = 60°

Q35. If sin(A - B) = sin A cos B - cos A sin B and cos (A - B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15.

Solution:

Given,

sin(A - B) = sin A cos B - cos A sin B

And, cos (A - B) = cos A cos B + sin A sin B

We need to find, sin 15 and cos 15.

Let A = 45 and B = 30

sin 15 = sin (45- 30) = sin 45 cos 30 - cos 45 sin 30

= = ( 7 2 X X_ _ _ _ _ _ V3-1=(1V2XV32)-(1V2X12)=V3-12V2—

cos 15 = cos (45- 30) = cos 45 cos 30 - sin 45 sin 30

= = ( ^ 2 X ^ + X

=(1V2XV32)+(1V2X12)=V3+12V2=

Q36. In a right triangle ABC, right angled at C, if ZB=60°ZP = 60° and AB=15 units. Find the remaining angles and sides.

Solution:

S in60°=x15 V32 = x15X=15V32UnitSCOS60°=x15 12 = x15X =152X =7.5un itS

sm60° = f5V3 _ x2 ~ 152 15

X

cosQO0 = §___ X

2 15

x = 7.5units

Q37. In AAABC is a right triangle such that Z C = 9 0 °ZC = 90°, Z A = 4 5 °ZA = 45° and BC = 7 units. Find the remaining angles and sides.

Solution:

C

B

7

A

X

Here, zC = 9 0°Z C = 90° and zA =45 °Z A = 45°

We know that,

z A + z B + z C Z A + Z B + Z C = 180°180°

=> 45°45° + 90°90° + z C Z C = 180°180°

=> 135°135° + z C Z C = 180°180°

=> z C Z C = 180°180° - 135°135°

=> Z.CZC = 45°45°

The value of the remaining angle C is 45°45°

Now, we need to find the sides x and y

here,

cos(45) = BCAB

1V2 -h= = 7 y -y/2 y y

y = 7V27\/2 units

sin (45) = ACAB

lV2 -i= = xy —2/

1V2 —7= = x7V2V2

x7 2

x = 7’& 5 ^

x = 7 units

the value of x = 7 units and y = 7V27-\/2 units

Q 38 . In a rectangle ABCD , AB = 20 cm , ZZBAC = 60°60° , calculate side BC and diagonals AC and BD .

Solution:

Let AC = x cm and CB = y cm

Since , COSQcosO = basehypotenuse t— —hypotenuse

Therefore , COs60°=20xcos60° = —

=> 12 —20x=> ^ [since,COs60°=12cos60° = -|]

=>=> x= 40 cm = AC

Similarly BD = 40 cm

Now,

Since , SinGsmf? = perpendicularhypotenuse PeJ'Pen icularhypotenuse

Therefore , sin60°=BCACsm60° = A C

=>V32 = y40=^ ^ ^ =>y=40V32^> y = ^

=>y=20V3^> y = 20a/3 cm .

Q39:lf A & B are acute angles such that tanA=1/2 tanB=1/3 and tan(A+B)=tanA+tanB1-tanAtanB tanA+tanB

1—tanA tanB, find A+B.

Solution:

1_|_I i± l 1T an(A+B)= 12 + 131—12.13 Tan(A + B ) — 2 1 31 3+2656 = —|---- 5656= -jr-

2' 3 6 6Tan(A+B)=56><65Tan{A + B) = § x § (A+B)=TarT1(1)(A + B) = T a n ^ i l )

(A+B) = 45°

Q40: Prove that: (V3-1)(3-C O t30o)=tan360-2s in60°( V 3 — 1)(3 — cot30°) = tan3 60 — 2sm60°

oOCO-wou

(0c<

ooCO"o

AIICOI

+ICO

' • 9

oCOo0

1CO

Jco

rH+

ICO

I co

+ +\co \co

<M

II

I

= 2V

32\/3

R.H

.S =

> ta

n360

-2si

n60°

£an3

60 —

2sm

60'o

XCM

'9xCMIco

I

= 3V

3-V

33v/

3 -

y/S

JPOCM

' 9CM

COIofIICOI

TJ(U>o

(U0 c (U1

/ /


Q.1) Evaluate the following : (i) sin20cos70

Sol ( i) : Given that, sin20cos70' ’ ’ cos 70

Since sin (90 - 0 0 ) = cos0@

sir, on sini90—70)=>=> sin20cos70-^|^= sin(90-70)cos70— 7n

=>=>- sin20cos70 sm 20 _cosjq ~ cos70cos70 cos70

cos70

=>=>- sin20cos70 sin2 0cos70 =1

Therefore sin20cos70 stnl® = 1cos 70

(ii) cos19sin71 cos!9sin71

Soln.(ii): Given that, cosl9sin7l cos!9sin71

=>=>- cos19sin71 cos\9 _ sin71

cos( 90—71)cos(90-71)sin71 sin71

=>=£>cos19sin71 cos!9sin71

= sin71sin71 sm71sin71

cos19sin71 cos!9sin71 = 1

Since cos(9O-0@)= sin 0 0

Therefore cosl9sin7l cos^ = 1

(Mi) sin21cos69 sin21cos69

Soln.(iii): Given that, sin2lcos69 sin21co«69

Since (90-00) = cos00

=>=y>sin21cos69 ^ gQ = sin(90-69)cos69

sin21cos69 sin 21 cos 69

= cos69cos69 cos 69 cos 69

sin21cos69 sin21cos69 =1

(iv) tan10cot80 tanlOcot80

Soln.(iv): We are given that, tanl0cot80 tanlOcot80

Since tan(9O-00) = cot00

-tan10cot80-^^ =tan(90-80)cot80 ian(90—80) cot80

tan10cot80

cot80cot80 cot80cot80

tanlO _ cot80

=>=>- tan10cot80 .on = 1cotm

T h e r e fo r e tan l0co t80tanlOcotSO = 1

(v) sec11cosec79 sec' ' cosecTd

Soln.(v):

Given that, sec11cosec79

Since sec(9O-O0)=cosec©0

=>=£> sed 1 cosec79 secllcosec79

sec(90-79)cosec79 sec(90—79) cosec79

=>=>- sed 1 cosec79 secllcosec79 = cosec79cosec79 cosec79

cosec79

=>=>- sed 1 cosec79 secllcosec79 = 1

Therefore sec11cosec79 aeclL = 1cosec79

Q.2: EVALUATE THE FOLLOWING :

(i) (sin49°cos41° \ / sin49° \ • V c o s4 1 °y

2+ (cos41°sin49° \ ( cos41° \

• \ sm 49° )

2

Soln.(i):

2 / . Q \ 2 2 / o \ 2We have to find: (sin49°cos4l°) ( ) + (cos4l°sin49°) ( cost\l )

v ' y cos41 ) v ' \ sin 4 9 ° J

Since sec70°cosec20° sec7®° + sin59°cos31° sin^ ° sin(90°90° - 0 0 ) = cos©@ and cos(90°COo g C ZU COStj J.

90° - 0 0 ) = sin0©

So

( sin(90°-41 °)cos41 ° ) ( ) + (cos(90°-49°)sin49°) ( )

( cos41°cos41°) + ( Sin49°Sin49°) ( g ^ )

= 1+ 1 = 2

So value of (sin49°cos41°) + (cos41°sin49°) is 2

(ii) cos48°cos48° - sin42°sm42°

Soln.(ii)

We have to find: COs48°cos48° - Sin42°sm42°

Since cos(9O°-09O° — 0 ) = sin00.So

cos48°cos48° -sin42°sm42° = cos(90°-420)-sin42°cos(900 - 42°) - sm42°

=sin42°sm42° - sin42°sin42° =0

So value of cos48°cos48° - Sin42°sm42° is 0

(iii) c o l4 0 - te n 5 0 -1 2 (c o ,3 5 -s in 5 5 - )^ “ l ( ^ § )

Soln.(iii)

We have to find:

cot40°tan50°— 12 ( cos35°sin55°) \ ( CQS )' ' tanoO i \ stnoo )

Since cot(9O°-09O° — 0 ) = tan0@ and cos(9O°-09O° — @)=sin©@

cot40°tan50°— 12 ( cos35°sin55°) \ ( ^*55° ) = cot(90°-50°)tan50° - 12(cos(90°-55°)sin55°)

coi(90°—50°)t c m 5 0 °

1 / cos(90°—55°) \2 \ sm 55° )

= tan50°tan50° -1 2 ( sin55°sin55°) 1. / sm 55° \ 2 ^ sm 55° y

= 1-121 - \ = 12±

So value Of C o t4 0 -ta n 5 0 -1 2 (co s 3 5 -s in 5 5 -)^ - l ( ^ g p ) is 12±

(iv)(sin27-cos63')2 ( ^ P ) 2 - (cos63-sln2r)2 ( t § - )

Soln(iv)

We have to find: (sin2Tcos63-)2( ^ ) 2 - (c o s e a w ^ ^ ) 2

Since sin (9O°-09O° — 0)= cos00 and cos (90°-©90° — 0)=sinO0

2 2 2 2 2 (sin27“cos63°) - (cos63'sin2r) = ( sin(90'-63’ )cos63-)

- ( ooS(90.-27> „ 2^ 2( ^ M ) 2

= ( cos63-oos63-)2 ( ^ | ; ) 2 _ ( sln2rsln2r )2 ( ^ ) 2 = 1-1 =0

Sovalueof(sin27-coS63-)2 ( ^ ) 2 -(cos63-sin2r ) 2 ( ^ | ; ) 2 is 0

(v)tan35°cot550 + co t7 8 °ta n 1 2 °-1 ^ !^ - +' ' cot55 tanl2 — 1

Soln.(v)

We have to find:

tan35°cot55° + cot78°tan12°— 1 ^ 77^ + f°*78° - 1cot55 tanl2°

/ sin(90°—63°) Y cos63°

Since tan (9O°-09O° — ©)= cot0© and cot (90°-©90° — 0)=tanO0 =1

So value of tan35oC0t55° + C0t78otan12°is1 + f°*7f 0 251catbo tanl2

(vi) sec70°cosec20° + sin59°cos31° sec no ' 7 cosec20 + sin59°

cos31°

Soln.(vi)

We have to find: sec70°cosec200 + sin590cos31° + sm 59°cos31°

Since sec70°cosec20° + sin59°cos310 sec7 0 + and sec(90°—090° — @)=cosec©@cosec20 cosZl ' '

So

2 2sec70°cosec20° + sin59°cos31 ° aec7°° + =( sec(90°-20°)cosec20°) f SeC ° ~^0° ) -

cosec20 cos31° ' 7 \ cosec20° /

( * „ ( 9 0 - - ~ 0 2( ^ ^ ) 2

= cosec200cosec200 + cos310cos31°-£2^ ^ + CQS ° =1+1 =2cosec20 ' cos31

So value of sec700cosec20° + sin590cos31° sec7°° + sm ° is 2COocCZU COdOl

(vii) cosec310-sec590cosec3l0 — sec59°.

Soln(vii)

We have to find: COSec31°-sec59°cosec3l° — sec59°

Since COSec(9O°-0)cosec(9O° — 0)=sec00.So

= cosec31°-sec590cosec3l° - sec59°

= cosec(90°-59°)-sec59cosec (90° - 59°) - sec59 =sec59°-sec59° sec59° - sec59° =0

So value of cosec31°-sec590cosec3l° — sec59° is 0

(viii)(sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° — cosl8°)

Soln.(viii)

We have to find: (sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° — cosl8°)

Since sin(9O°-0)sm(9O° — 0)=cos0©,So

(sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° - cosl8°)= (sin72°)2 (sin72°) 2-(cos 18°)2 (cos 18°)2

=[sin(90°-1 8°)]2- ( cos1 8°)2[sm (90° - 18°)]2 - (cosl8°)2

=(cos1 8°)2(cos18°)2- (cos1 8°)2(cos18°)2

=cos218°-cos218°cos218° - cos218° =0

So value of (sin72°+cos18°)(sm72° + cosl8°) (sin72°-cos18°)(sm72° — cosl8°) is 0.

(ix)sin35°sin550sm350sm550-cos350cos550cos350cos55°

Soln(ix)

We find :

Sin350sin550sm350sm550-cos350cos55°cos350cos550

Since sin(9O°-0)sm(9O° — 0)=cos00 and COS(9O°-0)cos(9O° — 0)=sinO0

Sin350sin550sm350sm550-cos350cos550cos350cos55°= sin(90°-55°)sin55° sm(90° — 55°)sm550-cos(900-550)cos550cos(90° — 55°)cos55° =1-1 =0

So value of sin35°sin550sm350sm550-cos350cos550cos350cos55° is 0

(x) tan480tan230tan420tan670£an48°tcm230£an420£an670

Soln.(x)

We have to find tan48°tan230tan420tan670tan480£an230£an420£an67°

Since tan(90°-©)£an(90° — @)=cot©@. So

tan480tan230tan420tan670£an480£an230tara42°£cm670 = tan(90°-42o)tan(90o-67o)tan42otan67°tan (90° - 42°) tan (90° - 67°) tan42°tan67°

=C0t42°C0t67otan42otan67°coM2occtf67o£an42o££m67o

=(tan670cot670)(tan420cot420)(£an67°co£670) (£an42°cot42°) =1*1 =1

So value of tan480tan230tan420tan670£an480£<m23°£cm420tan670 is 1

(xi)sec50osin40o+cos40ocosec50osec50osm40° + cos40°cosec50°

Soln.(xi)

We have to find sec500sin400+cos400cosec50°sec500sm400 + cos40°cosec50°

Since COS(9O°-0)cos(9O° — @)=sin0@,sec(9O°-0)sec(9O0 — 0)=cosec0© and sin00.cosec©0=1. So

sec50°sin400+cos400cosec500sec50°sm400 4- cos40°cosec50° = sec(90°-40o)sin40°sec(90o - 40°)sm40° +cos(90°-50°)cosec50° cos (90° — 50°)cosec50° =1+1 =2

So value of sec50osin40o+cos40ocosec50osec50°sm400 + cos40°cosec50° is 2.

Q.3) Express cos75°750+cot75°750 in terms of angle between 0° and 30°.

Soln. 3 :

Given that: cos75°750+cot75°750

=cos75075°+cot75°750

= cos(90°-15°)+cot(90°-15°)cos (90° - 15°) + cot (90° - 15°)

=sin 15°l5°+tan 15°15°

Hence the correct answer is sin 15°15°+tan 15°15°

Q.4) If sin3A = cos(A - 26°), where 3A is an acute angle, find the value of A.

Soln.4:

We are given 3A is an acute angle

We have: sin3A=cos(A-26°26°)

=>=^sin3A=sin(90°90o-(A-26o26°))

=>=>- sin3A=sin(116°116°-A)

=>= 3A=116°116°-A

=>=^4A=116°1160

=>=^A=29°29°

Hence thecorrect answer is 29°29°

Q.5)lf A, B, C are the interior angles of a triangle ABC, prove that,

(i) tan(c+A2)=cotB2tan = cot^

(ii) sin(B+c2)=cosA2sin = cos^

Soln.5:

(i)We have to prove: tan(c+A2)=COtB2tcm ( ^ y ^ ) = cot-f

Since we know that in triangle ABC

A+B+C=180

=>=>C+A=180°180°-B

=> > C+A2 —90°— B2 ~ = 90° - f

=>^tanc+A2=tan(90°B2)£an^^ = tan (90°-j)

=>=^tan(c+A2)=cotB2*an = cotf

Hence proved

(ii)We have to prove : sin(B+C2)=COSA2sin —

Since we know that in triangle ABC

A+B+C=180

=>=>B+C=180°180°-A

B+C2 —90°—A2 — = 90° - 4=>^sinB+C2=sin(90oA 2 )s m ^ 1 = sin (90°

sin(B+c2)=cosA2sm = cos4

Hence proved

Q.6)Prove that:

(i)tan20o20otan35o35otan45o45otan55o55otan70°70'

(ii) sin48°48° .sec48°48° +cos48°48° .cosec42°42° =

(iii) sin70°cos20° + cosec20°sec70°~2COS70°.COSGC20°:

+ coseg01 _ 2 co« 7 0 ° . cosec 20° = 0cos20 sec70

C O S

= 1

0

Soln.6:

(i) Therefore

tan20o20otan35o35otan45o45otan55o55otan70°70o

=tan(90°-70°)tan (90°-70°) tan(90°-55°)tan (90°-55°) tan45°45°tan55° 55°tan70°70°

=cot700700cot550550tan45°450tan550550tan700700

= (tan700cot700)(tan550cot550)tan450(£an700co£700) (tan&b0cotbb0) tan45° =1x1x1 =1

Hence proved

(ii) We will simplify the left hand side

sin48°48° .sec48°48° +cos48°48° .cosec42°42° =sin48°48°. sec(90°-48°) sec (90° - 48°)cos48°480 .cosec(90°-48°)cosec (90° - 48°)

=sin480480.cos480480+cos480480.sin48048° =1+1 =2

Hence proved

(iii) We have,sin70°cos20° + cosec20°sec70°—2cos70°.COSec20°=0

singl + _ 2cos70°. cosec20° = 0cos20 sec70

So we will calculate left hand side

sin70°cos20° + cosec20°sec70° -2cOS70°.COSec20°=0

(iv) c o s 8 0°s in io °+ c o s 5 9 °.c o s e c 3 1 °= 2 ^ g ^ + cos59°.cosec31° = 2

svn^ + cose^_ _ 2cos70°. cosec20° = 0=cos20 sec70

sin70°cos20° + cos70°sin20° -2COS7 0°. COSec(90°-70°)

sin70° cosec20°sec70°

- - 2cos70°.cosec(90° - 70°)cos 20' sin20

= sin(90°-20°)cos20'sm (90°— 20°)

cos20°

2cos70°.cosec(90° -7 0 °)

+ cos(90°-20°)si n20'cos(90°— 20°)

sin20° -2cos70°.cosec(90°-70°)

= cos20°cos20° + sin20°sin20°—2*1 - 2 X 1 =1+1-2 =2-2 =0

Hence proved

(iv)We have cos80°sinio°+cos590.cosec310= 2 ^ f^ - + cos59°. cosec31° = 2sm l0 °

We will simplify the left hand side

cos80°sin10° +cos590.cosec31°-^f^ + cos59°. cosec31°=sznlO

cos(90o-ioo)sinio°+cos590.cosec(900-590) cos(s9.°l0!° + cos59°. cosec(90° — 59°)

=sinioosinioo+cos590.sec59°^i^- + cos59°.sec59° =1+1 =2s*nlO

Hence proved.

Question 7

If A,B>C are the interior of triangle ABC , show that

(i) sin(B+c2)=cosA2sin(^±^) = cos 4

Solution

A+B+C=180°

B + C= 180°-A2 4

LHS=RHS

(ii)cos(90°-A2)=sinA2cos(90° - 4 ) = sin4

LHS=RHS

Question 8

If 20+45°and3O-©20 + 45° and 30 — 0 are acute angles , find the degree measure of

0 0 satisfying sin(2O+45o)=COS(3O°+0)sin(2O + 45°) = cos(30° + 0)

Solution

Here 2O+45°=sin(6O°+0)sin(6O0 + 0)

We know that ,((9O°-0)(9O° — 0)= cos(0)cos(0)

= sin(20+45°)=sin(9Oo-(3O°-0))sin(20 + 45°) = sin(90° - (30° - 0 ))

= sin(20+45o)=sin(9Oo-3O°-0)sin(2© + 45°) = sin(90° - 30° - 0)

= sin(20+45o)=sin(6O°+0)sin(2© + 45°) = sin(60° + 0)

On equating sin of angle of we get,

= 2©+45°=6O°+©20 + 45° = 60° + 0 = 0=15°0 = 15°

Question 9

If 0 0 is appositive acute angle such that sec0=CSc6O°sec 0 = esc 60° , find

2 c o s 20 - 1 2 cos2 0 — 1

Solution

We know that, sec(9O°-0)=CSC20sec(9O° — 0) = esc2 0

= sec(©)=sec(90°-60°)sec(©) — sec(90° - 60°)

= ©=3O°0 = 30° = 2cos2©-1 2 cos2 0 - 1

= 2cos230-1 2 cos2 30 - 1

= 2(^32)2- 1 2 ( ^ ) 2 - 1

= 2(34)-12(|) - 1 =(32)-1(f) - 1 = (12)(|)

Qio.if sin30=cos(0-6°)where30and0-6circsin30 = cos(@ — 6°) where 30 and 0 — 6c*rc acute angles, find the value of 0 0 .

Soln:

We have, sin30=cos(0-6°)sin3© = cos(© — 6°)

cos(9O°+30)=cos(0-6°)cos(9O0 + 30) = cos(0 — 6°) 9O°-30=0-6°9O° 30 = 0 -6 ° -30-0=6°-9O°—30 - 0 = 6° - 90° -40=96°-4© = 96° O=-96°-4=24o0 = ^ = 24°

Q11 .If sec2A=CSC(A-42°)sec 2A = csc(A — 42°) where 2A is acute angle, find the value of A.

Soln:we know that sec(9O-3Theta)=CSC0sec (90 — 3 Theta) = esc©

sec2A=sec(90-(A-42))sec 2A = sec(90- (A - 42)) sec2A=sec(90-A+42)) sec2.A = sec(90-24 + 42)) sec2A=sec(132-A))sec2A = sec(132-.A))

Now equating both the angles we get

2A = 132 - A

A = =1323= ^

A = 44

exercise 5.1: trigonometric ratios - kopykitab · exercise 5.1: trigonometric ratios 1.) find the...

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