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.

(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

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

Therefore,

By Pythagoras theorem,

AC2 = AB2 + BC2

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

52 = 42+ BC2

BC2 = 52 - 42

BC2= 2 5 -1 6

BC2 = 9

Hence, Perpendicular side = 3

■ * • « 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

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

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

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

(i) The given triangle is below:

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

To find, tan P, cot R

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.

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

By definition,

COSA= BasesideadjacenttozAHypotenuse COS A

COS A = COSA= 4041 COS A =

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

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.

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

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

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

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

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

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

Therefore,

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

X sin 02 cos 0 + s in 0

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

Given:

3cotO=23 cot 0 = 2

Therefore,

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

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

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

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

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

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

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

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 ) — ^

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

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,

Given: COt©= W3cot @ ..(1)

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

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

Therefore,

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

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

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

5 6 + 8

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

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,

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 ©

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

312_ 313- 31231325313 - j j "

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

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,

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

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

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,

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

JS__ 15- -815 83 —

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

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.

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

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

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

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

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

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

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.

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

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 + |

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.

(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

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

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

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

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

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

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

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 (# ) '

- 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 -61164 — ^

= 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

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

Now we know that

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

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

To find: tanOtan 0

We can write this as:

Dividing both the sides by COS0cos 0 ,

We get,

Hence,

tanO=1tan 0 = 1

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

zA = zP Z .A = Z P

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

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°

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| + \ + \

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

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°)

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 )

(O09soo

+Osfrsoo-O

06u!s)(0oeu!s+Osfru!s+Oosoo) c

05COCO

■ss| h

t o I CO

CO |N3

^N>Xco

N> IN3

455.XN3

to| I—.to

CD II Q W CDOct> o < o O

to o CD c toCO o COo

CO 5' oo

tocdoCo

>HcoCoCD<\oOr01

CoCD<\o03O

IN>OO</)N305O

toCDOCo

1 1CO £x|CO</) CoCD CD°N 3455.Olo1

1- kC/5 CoCD CDO

N3COoo

hOOOr+-toCO

IS)oowto05

coWCD0to01oI

wCDOtoCO

u N3t o

(SA) x

(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°

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

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

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°

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

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

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^ \

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.

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

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

cosQO0 = §___ X

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:

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

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

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

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

ooCO"o

AIICOI

' • 9

+ +\co \co

60 —

'9xCMIco

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

=>=>- 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

=>=£> 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° )

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

( 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 )

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

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°

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°

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°)

(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°

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°

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

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

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 ))

On equating sin of angle of we get,

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

= 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 .

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.

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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