2....1. express the trigonometric ratios sinθ , sec θ and tan θ in terms of cot θ. 2. write all...
TRANSCRIPT
Introduction to Trigonometry
317
NCERT Textual Exercises and Assignm
entsExercise – 7.1
1. In ∆ABC,right-angledatB, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A, (ii) sin C, cos C. 2. IntheFig.7.7,findtanP – cot R.
Fig. 7.7
3. If sin θ = 34,calculatecosθ and tan θ.
4. Giventhat15cotθ = 8. Find sin θ and sec θ.
5. If sec θ = 1312
,calculateallothertrigonometricratios.
6. If ∠A and ∠BareacuteanglessuchthatcosA = cos B,provethat∠A = ∠B.
7. If cot θ = 78 .Evaluate.
(i) ( sin ) ( sin )( cos ) ( cos )1 11 1+ −+ −
θ θθ θ
(ii) cot2 θ
8. If 3 cot θ=4,checkwhether 11
2
2 −+
tantan
θθ
= cos2 θ – sin2 θ or not.
9. IntriangleABC,rightangledatB, if tan A = 13,findthevalueof
(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C 10. In ∆PQR,rightangledatQ, PR + QR = 25 cm and PQ = 5 cm. Determine
thevalueofsinP, cos P and tan P. 11. Statewhetherthefollowingstatementsaretrueorfalse.Justifyyouranswer. (i) Thevalueoftanθisalwayslessthan1.
(ii) sec θ = 125
forsomevalueofangleθ.
(iii) cos θistheabbreviationusedforcosecantofangleθ.
Introduction to Trigonometry
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NCERT Textual Exercises and Assignm
ents (iv) cotθ is the product of cot and θ.
(v) sinθ = 43forsomeangleθ.
Test Yourself – TRG 1
1. GiventhattanA = 43.FindtheothertrigonometricratiosofangleA.
2. GiventhatsinA = 35.FindtheothertrigonometricratiosofangleA.
3. Consider ∆ACB,rightangledatC, in which AB = 29 units, BC = 21 units and ∠ABC = θ.Determinethevalueof
(i) cos2 θ + sin2 θ (ii) cos2 θ – sin2 θ 4. InatriangleABC,rightangledatB, if tan A=1,verify2sinA.cos A = 1. 5. In ∆OPQ,rightangledatP, OP = 7 cm and OQ – PQ = 1 cm. Determine
thevalueofsin Q and cos Q. 6. If ∠B and ∠QareacuteanglessuchthatsinB = sin Q,provethat∠B = ∠Q.
7. If 3 cot θ=2,thevalueof 2sin ¸ 3cos¸2sin ¸ +3cos¸
− .
8. If sin θ = 35,thevalueof(tanθ + sec θ).
9. If tan θ + cot θ =2,findthevalueoftan²θ + cot² θ .
10. If 4 tan A=3,find4 24 3
12 2
2 2
sin cossin cos
A AA A−+
+ .
Exercise – 7.2 1. Evaluatethefollowing. (i) sin600cos300+sin300cos600
(ii) 2 tan2 450 + cos2300 – sin2600
(iii) cos
sec
cosec 45
30 30
0
0 0+
(iv)sin tansec cos cot
cosec 30 45 6030 60 45
0 0 0
0 0 0
+ −+ +
Introduction to Trigonometry
319
NCERT Textual Exercises and Assignm
ents (v) 5 60 4 30 45
30 30
2 0 2 0 2 0
2 0 2 0
cos sec tansin cos
+ −+
2. Choosethecorrectoptionandjustifyyourchoice.
(i) 2 301 30
0
2 0
tantan
+
=
(A) sin600 (B) cos600
(C) tan600 (D) sin300
(ii) 1 451 45
2 0
2 0
−+
tantan
=
(A) tan900 (B) 1 (C) sin 450 (D) 0 (iii) sin2A=2sinAistruewhenA= (A) 00 (B) 300
(C) 450 (D) 600
(iv)2 30
1 30
0
2 0tantan
−
=
(A) cos600 (B) sin600
(C) tan600 (D) sin300
3. If tan (A + B) = 3 and tan (A – B) = 13,00 < A + B < 900, A > B,findA
and B. 4. Statewhetherthefollowingstatementsaretrueorfalse.Justifyyouranswer. (i) sin (A + B) = sin A + sin B (ii) Thevalueofsinθ increases as θ increases. (iii) Thevalueofcosθ increases as θ increases. (iv) sinθ = cos θ forallvaluesofθ. (v) cotAisnotdefinedforA=0º.
Test Yourself – TRG 2 1. In ∆ABC,rightangledatB, AB = 5 cm, ∠ACB=30º.Determinethelength
of the sides BC and AC. 2. In ∆PQRrightangledatQ, PQ = 3 cm and PR = 6 cm. Determine ∠QPR
and ∠PRQ.
Introduction to Trigonometry
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NCERT Textual Exercises and Assignm
ents 3. If sin (A – B) = 1
2, cos (A + B) = 1
2,00 < A + B <900, A > B,findA and B.
4. Evaluatecos²300 cos² 450+4sec²600 + 12cos²900–2tan²600.
5. If ∠A=300,verifythatsin2A = 21 2
tantan
AA+
.
6. If ∠A = ∠B=45º,verifythatsin(A + B) = sin A cos B + cos A sin B.
7. Evaluatesin cos tan
cot cos tan
² ²
² ²
30 2 45 6012
45 30 45
0 0 0
0 0 0
+ +
+ +.
8. Evaluate 13
30 13
60 12
45 23
602 2 2 2tan cos sec sin ° − ° + ° − °
9. If ∠A=60°,verifythat12 2 2
sin sin .cosA A A=
10. IftanA= 12andtanB= 1
3,find(A+B)byusingthefollowingformula
tan(A+B)=tan tan
tan .tanA BA B+
−1 .
Exercise – 7.3 1. Evaluate:
(i) sincos
1872
0
0 (ii) tancot
2664
0
0
(iii) cos 480 – sin 420 (iv) cosec310 – sec 590
2. Show that: (i) tan 480 tan 230 tan 420 tan 670 = 1 (ii) cos 380 cos 520 – sin 380 sin 520=0 3. If tan 2A = cot (A – 180), where 2Aisanacuteangle,findthevalueofA. 4. If tan A = cot B,provethatA + B=900. 5. If sec 4A = cosec (A–200), where 4Aisanacuteangle,findthevalueofA. 6. If A, B and C are the interior angles of a triangleABC, show that
sin B C A +
=2 2
cos .
7. Expresssin670 + cos 750intermsoftrigonometricratiosofanglesbetween
Introduction to Trigonometry
321
NCERT Textual Exercises and Assignm
ents00and 450.
Test Yourself – TRG 3 1. Evaluate:
(i) sincos
4941
0
0 (ii) tancot
5931
0
0
(iii) sec cosec
3555
0
0 (iv) cos700–sin200
(v)sincos
cossin
4941
4149
0
0
2 0
0
2
+
(vi)
sincos
cossin
2763
6327
0
0
2 0
0
2
−
(vii)cossin
cossin
4050
12
3555
0
0
0
0−
(viii)
sin ºcos º
tan ºcot º
2070
12
2070
+
(ix)tancot
seccos
1872
783
0
0
0
0−ec
(x) cos cossin sin
2 0 2 0
2 0 2 0
20 7059 31
++
(xi) cossin
700
020 + cos 690 cosec 210
2. Provethat (i) sin 630 cos 270 + cos 630 sin 270 = 1 (ii) sin 480 sec 420 + cos 480 cosec 420 = 2
(iii) sec
csincos
osec
3753
4248
0
0
0
0+ = 2
(iv) tan50tan100 tan 150 tan 750tan800 tan 850 = 1
(v)sin sin sincos cos cos
10 20 3080 70 60
0 0 0
0 0 0 = 1
(vi)cossin
cossin
8010
5931
0
0
0
0+ = 2
(vii) sin2 350 + sin2 550 = 1 (viii) sec500sin400+cos400cosec500 = 2 (ix) cosec2 740 – tan2 160 = 1 (x) sec2 120 – cot2 780 = 12
Introduction to Trigonometry
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NCERT Textual Exercises and Assignm
ents (xi)
cottan
tancot
5436
2070
20
0
0
0+ − =0
(xii)cossin
cossin
sin
7020
5931
8 300
0
0
02 0+ − =0
(xiii)sincos
cossin
cos
4743
4347
4 450
0
2 0
0
22 0
+
− =0
(xiv) sec700sin200–cos200cosec700=0 (xv) sin2200 + sin2700 – tan2 450=0 3. Expresseachoneofthefollowingintermsoftrigonometricratiosofangles
lyingbetween00 and 450. (i) cos 780 + tan 780 (ii) sin 840 + sec 840
(iii) cos 560 + cot 560 (iv) sin850 + cosec 850
(v) cosec690 + cot 690
4. Expresscos750 + cot 750intermofanglesbetween00and300. 5. Provethat
(i) sin cos cos ( )
sin ( )sin cos sin ( )
cos ( )θ θ θ
θθ θ θ
θ90
9090
90
0
0
0
0
−−
+−
−= 11
(ii) sin(900–A)cosA+cos(900–A)sinA=1
(iii) sin(900 – α)cos(900 – α) =tan
tanαα1 2+
(iv) cossin (90 )
+ sincos (90 )
= 20 0
θθ
θθ− −
(v) sinθsin(900 – θ) – cos θ cos(900 – θ)=0 (vi) cosθ sin(900 – θ) + sin θcos(900 – θ) = 1
Exercise – 7.4 1. Expressthetrigonometricratiossinθ, sec θ and tan θ in terms of cot θ. 2. Writealltheothertrigonometricratiosof∠A in terms of sec A. 3. Evaluate
(i) sin ² sin ²cos ² cos ²
63 2717 73
0 0
0 0
++
(ii) sin 250 cos 650 + cos 250 sin 650
4. Choosethecorrectoption.Justifyyourchoice. (i) 9 sec2 A – 9 tan2 A (a) 1 (b) 9
Introduction to Trigonometry
323
NCERT Textual Exercises and Assignm
ents (c) 8 (d) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (a) 0 (b) 1 (c) 2 (d) 0 (iii) (sec A + tan A) (1 – sin A) (a) sec A (b) sinA (c) cosec A (d) cos A
(iv)11
2
2
++
tancot
AA
(a) sec2 A (b) –1 (c) cot2 A (d) tan2 A 5. Provethefollowingidentitieswheretheanglesinvolvedareacuteangles
forwhichtheexpressionsaredefined.
(i) ( cot ) coscos
cosec
θ θθθ
− =−+
2 11
(ii) cos
sinsin
cosA
AA
A11
++
+= 2 sec A
(iii) tan
1 cot+ cot
1 tanθθ
θθ− −
= 1 + sec θ. cosec θ.
(iv) 11
2+=
−sec
secsin
cos
A
AA
A
(v)cos sincos sin
A AA A− ++ −
11
= cosec A + cot A
usingtheidentitycosec²A = 1 + cot² A.
(vi)11+−
sinsin
AA
= sec A + tan A
(vii) sin sincos cos
tanθ θθ θ
θ−−
=2
2
3
3
(viii) (sinA + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
(ix) (cosecA – sin A) (sec A – cos A) = 1
tan cotA A+
(x)11
11
2
2
22+
+=
−−
=
tancot
tancot
tanAA
AA
A
Introduction to Trigonometry
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NCERT Textual Exercises and Assignm
entsTest Yourself – TRG 4 Provethefollowingidentities: 1. sec A (1 – sin A) (sec A + tan A) = 1
2. cot coscot cos
A AA A
AA
−+
=−+
coseccosec
11
3. sin cos +1sin +cos 1
= 1sec tan
θ θθ θ θ θ−
− − usingtheidentitysec2 θ = 1 + tan2 θ.
4. cos + tan ¸ 1
sin = tan
2 2
22θ
θθ
−
5. tan4 A + tan2 A = sec4 A – sec2 A
6. tantan 1
+ cosecsec cosec
= 1sin cos
2
2
2
2 2 2 2
θθ
θθ θ θ θ− − −
7. cosecAcosecA 1
+ cosecAcosecA +1
= 2sec A2
−
8. sin + cossin cos
+ sin cossin +cos
= 21 2cos
= 222
θ θθ θ
θ θθ θ θ−−
− ssin 12θ − 9. (cosec A – sin A) (sec A – cos A) (tan A + cot A) =1 10. tan +cot +2=sec cosec2 2 2 2φ φ φ φ 11. tan² θ – sin² θ = tan² θ . sin² θ. 12. sin6 θ + cos6 θ = 1 – 3 sin² θ cos² θ 13. (1 + cot A – cosec A)(1+ tan A + sec A) = 2
14. 1 cos A1+ cos A
= sin A
1+cos A−
15. 2 sec² θ – sec4 θ – 2 cosec² θ + cosec4 θ = cot4 θ – tan4 θ
Introduction to Trigonometry
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NCERTTextualExercisesandAssignments
Exercise – 7.1 1. In ∆ABC,wehave ∴ (AC)2 = (AB)2 + (BC)2 [Pythagorastheorem] ∴ (AC)2 = (24)2 + (7)2
∴ (AC)2 = 576 + 49 ∴ AC = 625 ∴ AC = 25 cm Now,
(i) sin A BCAC
= =725
cos A ABAC
= =2425
(ii) sinC ABAC
= =2425
Fig. 7.8
cosC BCAC
= =725
2. In ∆PQR,usingPythagorastheorem, (PR)2 = (PQ)2 + (QR)2
∴ (13)2 = (12)2 + (QR)2
∴ 169 – 144 = QR2
∴ QR = 25 ∴ QR = 5cm
Now tan P QRPQ
= =5
12
cot R QRPQ
= =5
12 Fig. 7.9
∴ tan cotP R− = −5
125
12 ∴ tan P – cot R = 0
Introduction to Trigonometry
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3. sin θ = 34 .....(i) ...[Given]
sin θ = BCAC
....(ii) ...[Trigonometricratios]
∴ BCAC
=34 ...[from(i)and(ii)]
∴ Let BC = 3k,AC=4k ...[wherekisapositivenumber] In right ∆ABC,wehave ∴ AC2 = AB2 + BC2 ...(Pythagorastheorem) ∴ (4k)2 = AB2 + (3k)2
∴ 16k2 – 9k2 = AB2 A
B C
4k
3k
∴ 7k2 = AB2
∴ AB = 7 k .....(iii)
Now, cosθ = =ABAC
kk7
4
cos θ = 7
4
tan A BCAB
kk
= =37 Fig. 7.10
∴ tan θ = 37
4. Given:15cotθ = 8
∴ cot θ = 815 .....(i)
cot θ = ABBC .....(ii)
∴ 815
=ABBC
.....(iii)
∴ Let AB = 8k, BC = 15k...[wherekispositivenumber] Fig. 7.11 In right ∆ABC,usingPythagorastheorem, (AC)2 = (AB)2 + (BC)2
∴ (AC)2 = (8k)2 + (15k)2
∴ (AC)2 = 64k2 + 225k2
∴ AC2 = 289k² ∴ AC = 17k
sin θ = =BCAC
kk
1517
Introduction to Trigonometry
329
∴ sinθ = 1517
secθ = =ACAB
kk
178
∴ secθ = 178
5. secθ = 1312
.....(i) ...[Given]
secθ = ACAB .....(ii)
ACAB
=1312
...[from(i)and(ii)]
∴ Let AC = 13k, AB = 12k ...[wherekisapositivenumber] Now, In right ∆ABC, UsingPythagorastheorem, (AC)2 = (AB)2 + (BC)2
∴ (13k)2 = (12k)2 + (BC)2
169k2 = 144k2 + BC2
∴ 169k2 – 144k2 = BC2
∴ BC² = 25k²
∴ BC k= 25 2
BC = 5k Fig. 7.12
Now, sinθ = =BCAC
kk
513
∴ cosec θ = =ACBC
kk
135
∴ sinθ = 513
∴ cosec θ = 135
cosθ = =ABAC
kk
1213
cotθ = = =ABBC
kk
1213
125
∴ cotθ = 1213
∴ cotθ = 125
tanθ = =BCAB
kk
512
∴ tanθ = 512
Introduction to Trigonometry
330
6. ConsidertworighttrianglesAMN and BPQ in which ∠A and ∠B are acute and cos A=cosB
Now, cos A = AMAN
cos B = BPBQ
∴ AMAN
BPBQ
AMBP
ANBQ
= ⇒ =
Let AMBP
ANBQ
= = k(say) ....(i) Fig. 7.13 Fig. 7.14
∴ AM = kBP, AN = kBQ ∴ In right ∆AMN, In right ∆BPQ, AN² = AM² + MN² ....(Pythagorastheorem) BQ² = BP² + PQ² ....(Pythagorastheorem) ∴ (kBQ)² = (kBP)² + MN² ∴ PQ² = BQ² – BP² ∴ MN² = k²BQ² – k²BP² ∴ MN² = k² (BQ² – BP²)
Now, MNPQ
k BQ BPBQ BP
²²
² ² ²² ²
=−( )−( )
∴ MNPQ
k MNPQ
k²²
²= ⇒ = ..... (ii)
From (i) and (ii), we get
AMBP
ANBQ
MNPQ
= =
∴ ∆AMN ~ ∆BPQ .....(SSSsimilarity) ∴ ∠A = ∠B .....(correspondinganglesofsimilartriangles) AlternateMethod: Consider a right angled∆ABC where ∠C = 90° and∠A and ∠B are acute angles and
cos A = cos B.
Now, cos A ACAB
=
cosB BCAB
=
ItisgiventhatcosA = cos B
∴ ACAB
BCAB
=
⇒ AC = BC Fig. 7.15 ∴ In right ∆ACB, ∠A = ∠B
Introduction to Trigonometry
331
7. cotθ = =78
ABBC
Let AB = 7k, BC = 8k (where kisapositivenumber) In right ∆ABC,usingPythagorastheorem (AC)2 = (AB)2 + (BC)2
(AC)2 = (7k)2 + (8k)2 (AC)2 = 49k2 + 64k2 (AC)2 = 113k2
AC k= 113 2
AC k= 113
Now, sinθ = BCAC
=8113
kk
=8113 Fig. 7.16
and cotθ = ABAC
=7113
kk
=7113
(i) ( sin ) ( sin )( cos ) ( cos )
sincos
1 11 1
11
2
2+ −+ −
=−−
θ θθ θ
θθ
= −
÷ −
1 8113
1 7113
2 2
= −
÷ −
1 64
1131 49
113
=−
÷
−
113 64113
113 49113
= ×49
11311364
∴ ( sin )( sin )( cos )( cos )1 11 1
4964
++ −−++ −−
==θθ θθθθ θθ
Introduction to Trigonometry
332
(ii) cot227
8θ =
cot2 4964
θ =
8. 3 cot θ = 4
cot θ = =
43
ABBC
Let AB = 4k, BC = 3k (where kispositivenumber) In right ∆ABC,usingPythagorastheorem, (AC)2 = (AB)2 + (BC)2
(AC)2 = (4k)2 + (3k)2
(AC)2 = 16k2 + 9k2
(AC)2 = 25k2
AC k= 25 2 = 5k
sin θ = = =BCAC
kk
35
35 Fig. 7.17
cosθ = = =ABAC
kk
45
45
tanθ = = =BCAB
kk
34
34
LHS = 11
2
2−+
tantanθθ
= −
÷ +
1 34
1 34
2 2
= −
÷ +
1 9
161 9
16
16 9
1616 9
16−
÷
+
= ÷7
162516
= ×7
161625
=725
RHS = cos2 θ – sin2 θ
Introduction to Trigonometry
333
=
−
45
35
2 2
= −1625
925
=725
LHS = RHS Henceproved. 9. In ∆ABC,
tan A BC
AB= =
13
Let BC = 1k, AB = 3 k (where kisapositivenumber) In right ∆ABC,usingPythagorastheorem (AC)2 = (AB)2 + (BC)2
( ) ( ) ( )AC k k2 2 23 1= + (AC)2 = 3k2 + k2
(AC)2 = 4k2
AC k= 4 2 = 2k
sin A BC
ACkk
= = =12
12
cos A AB
ACkk
= = =3
23
2 Fig. 7.18
sinC AB
ACkk
= = =3
23
2
cosC BC
ACkk
= = =12
12
(i) sin A cos C + cos A sin C = ×
+ ×
12
12
32
32
= +14
34
=+1 34
=44
sin A cos C + cos A sin C = 1
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334
(ii) cos A cos C – sin A sin C = ×
− ×
32
12
12
32
= −3
43
4
=−3 34
=04=0
cos A cos C – sin A sin C = 0 10. In ∆PQR, PR + QR=25 [Given] Let QR = x ∴ x + PR = 25 ∴ PR = (25 – x) cm In right ∆PQR,usingPythagorastheorem (PR)2 = (PQ)2 + (QR)2
∴ (25 – x)2 = (5)2 + (x)2
∴ 625–50x + x2 = 25 + x2
∴ –50x = 25 – 625
∴ x = −−
=60050
12
∴ QR = 12cm, PR = 25 – x = 25 – 12 = 13 cm Fig. 7.19
sin P QRPR
=
∴ sin P =1213
cosP PQPR
=
cos P =5
13
tan P QRPQ
=
tan P =125
11. (i) FALSE.Sincetan600= 3 1> (ii) TRUE. Since sec θisalways≥ 1
Introduction to Trigonometry
335
(iii) FALSE,cosθisabbreviationusedforcosineθ. (iv) FALSE,cotismeaninglesswithoutanangleθ.
Here, cot side adjacent to
side opposite toθ
θθ
=∠∠
(v) FALSE,sinθisnevergreaterthan1.
Test Yourself – TRG 1
1. sin , cos , cot , , secA A A A A= = = = =45
35
34
54
54
cosec
2. cosec A A A A A= = = = =53
54
43
45
34
, cos , tan , sec , sec
3. (i) cos2 θ + sin2 θ = 1 (ii) cos2 θ – sin2 θ = 41
841
5. (ii) sin ,cosθ θ= =725
2425
7. 0 8. 4 9. 2 10. 22/21
Exercise – 7.2
1. (i) sin600cos300+sin300cos600 = 3
23
212
12
× + ×
= +34
14
=+3 14
=44
= 1
(ii) 2 tan2 450 + cos2300 – sin2600= 2 × (1)2 + 32
32
2 2
−
= + −2 34
34
= 2
(iii) cos
sec 45
30 3012
23
20
0 0+=
÷ +
cosec
=
÷
+
12
2 2 33
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336
= ×
÷
+×
12
22
2 2 33
33
(rationalisingthedenominator)
= ÷+2
22 3 6
3
= ×+
22
32 3 6
=+
3 24 3 12
=+
×−−
3 212 4 3
12 4 312 4 3
(rationalisingthedenominator)
=
−
− ( )3 2 12 4 3
12 4 32
( )
( )²
=−−
36 2 12 6144 48
)
=−12 3 2 6
96( )
=−3 2 68
(iv) sin tansec cos cot
30 45 6030 60 45
12
11
23
0 0 0
0 0 0
+ −+ +
= + −
cosec ÷ + +
23
12
11
= + − ×
÷ × + +
12
11
23
33
23
33
12
11
= −
÷
12
+ 11
2 33
2 33
+ 12
+ 11
=−
÷
3 + 6 4 36
4 3 + 3 + 66
=−
×9 4 3
66
4 3 + 9
=−9 4 3
9 + 4 3
=−
×−−
9 4 39 + 4 3
9 4 39 4 3
(rationalisingthedenominator)
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337
=−−
(9 4 3)(9) (4 3)
2
2 2
=−
−(9) 2 × 9 × 4 3 +(4 3)
81 (16 × 3)
2 2
=−
−81 72 3 + 48
81 48
=−129 72 333
=−3 (43 24 3)33
=−43 24 311
(v) 5 60 4 30 4530 30
5 12
4 23
2 0 2 0 2 0
2 0 2 0
2cos sec tansin cos
+ −+
= ×
+ ×
−
÷
+
22
2 2
1 12
32
( )
= −
÷
5 × 1
4+ 4 × 4
31 1
4+ 3
4
= −
÷
54
+ 163
1 1 + 341
=+ −
÷
15 64 1212
44
=−
×79 12
121
=6712
2. (i) LHS =+2 30
1 30
0
2 0
tantan
= ×
÷ +
2 13
1 13
2
= ÷ +
23
1 13
= ÷
23
3 + 13
Introduction to Trigonometry
338
= ÷23
43
= ×24
343
=×
33 2
= ×3
2 333 (rationalisingthedenominator)
=×
=3 33 2
32
RHS=sin600
=
32
LHS = RHS
2 30
1 30
0
2 0
tantan+ = sin 600 ⇒ Option(A)iscorrect.
(ii) LHS =−1 45
1 45
2 0
2 0
tan_ tan
=−+
1 11 1
2
2
( )( )
=−+
1 11 1
=02
=0
1 451 45
2 0
2 0
−+
tantan
= 0 ⇒ Option (D) is correct.
(iii) PutA=00 LHS = sin 2A =sin2×00
=sin00
=0 RHS = 2sin A =2×sin00 (sin00=0) =2×0 =0
Introduction to Trigonometry
339
sin 2A = 2 sin AistruewhenA = 00 ⇒Option(A)iscorrect.
(iv) LHS = 2 301 30
0
2 0
tantan−
= ×
÷ −
2 13
1 13
2
= ÷ −
23
1 13
= ÷−
23
3 13
= ÷23
23
= ×23
32
=33
= ×33
33 .....(rationalisingthedenominator)
= =3 3
33
=tan600 ..... (∴tan600 = 3 )
2 30
1 30
0
2 0
tantan− = tan 600 ⇒ Option (C) is correct.
3. tan A B+( ) = 3 ....(i) [given]
tan 60 3° = .... (ii) ∴ A + B=600 ....(iii) [From(i)and(ii)] tan (A – B) =
13 ....(iv) [given]
∴ tan300= 13 ....(v)
∴ A – B=300 ....(vi) [Fromequation(iv)and(v)] Addingequations(iii)and(vi),weget 2A=900
∴ A = 902
∴ ∠A = 450
Introduction to Trigonometry
340
SubstitutingA = 450 in equation (iii) we get 450 + B=600
∴ B=600 – 450
∴ ∠B = 150
4. (i) FALSE If A=600, B=300, then LHS=sin(60º+30º) =sin90º = 1 RHS = sin A + sin B =sin60º+sin30º
32
12
1 32
1+ =+
≠
∴ LHS ≠ RHS (ii) TRUE
sin00=0,sin300 = 12
0 5 45 12
22
1 412
= ° = = =. , sin . =0.7(approx.)
sin600 = 3
21 73
2=
.=0.87(approx.),sin900 = 1
i.e., Value of sin θ increases as θ increases from 00 to 900. (iii) FALSE
When cos , cos . . ( )0 1 30 32
1 732
0 870 0= = = = approx
cos . ( .), cos . , cos45 12
0 7 60 12
0 5 90 00 0 0= = = = =approx
i.e., Value of cos θdecreasesasθ increases from 00 to 900. (iv) FALSE
If θ =300 ⇒sin300 = 12
30 32
, cos ° =
i.e.,sin300 ≠cos300
Onlysin450 = cos 450= 12
(v) TRUE
cot cos
sin0 0
010
00
0= = = meaningless
∴ cot 00isnotdefined.
Introduction to Trigonometry
341
Test Yourself – TRG 2 1. AC=10cm,BC = 5 3 cm 2. ∠PRQ=300, ∠QPR=600
3. A = 450, B = 150 4. 83/8 7. 2 8. 19/36 10. 450
Exercise – 7.3
1. (i) sincos
cos( )cos
1872
90 1872
0
0
0 0
0=−
......[ sin θ=cos(90– θ)]
=cos 72cos 72
0
0
= 1
(ii) tan 26cot 64
0
0 =−cot ( )
cot90 26
64
0 0
0 ......[ tan θ=cot(90– θ)]
=cot 64cot 64
0
0
= 1 (iii) cos 480 – sin 420 =sin(900 – 480) – sin 420 ......[cos θ=sin(900 – θ)] = sin 420 – sin 420
= 0 (iv) cosec 310 – sec 590=sec(900 – 310) – sec 590 ......[cosec θ=sec(900 – θ)] = sec 590 – sec 590
= 0 2. (i) LHS = tan 480 tan 230 tan 420 tan 670
=cot(900 – 480)cot(900 – 230). tan 420 tan 670 ......[tan θ=cot(900 – θ)] = cot 420 cot 670 tan 420 tan 670
= 142
167
42 670 00 0
tan.tan
. tan . tan ...... cottan
θθ
=
1
= 1 = RHS LHS = RHS Henceproved. (ii) L.H.S. = cos 380 cos 520 – sin 380 . sin 520
=sin(900 – 380) . cos 520–cos(900 – 380). sin 520 ..........∴ −
−
sin = cos (90º )and cos = sin (90º )
θ θθ θ
= sin 520. cos 520– cos 520 . sin 520 =0
Introduction to Trigonometry
342
= RHS LHS = RHS Henceproved 3. tan 2A = cot (A – 180) ∴ cot(900 – 2A) = cot (A – 180) ........[ tan θ=cot(90º– θ)] ∴ 900 – 2A = A – 180
∴ –2A – A = – 180–900
∴ –3A=–1080
∴ A =°108
3 ∴ ∠A = 360
4. tan A = cot B cot(90–A) = cot B ..........[tanθ=cot(90º– θ)] 90º–A = B 90º=A + B A + B = 900
Henceproved. 5. sec 4A = cosec (A–200) ∴ cosec(90º–4A) = cosec (A–200) ..........[sec θ=cosec(90º– θ)] ∴ 90º–4A = A–20º ∴ – 4A – A=–20º–90º ∴ – 5A=–110º
∴ A =−−110
5 ∴ ∠A = 220
6. A + B + C=1800
∴ A B C+ +
=2
1802
0
..........[Dividingbothsidesby2]
∴ A B C+ += °
290
∴ B C A+
= ° −2
902
∴ sin sinB C A+
= −
2
902
0
∴ sin cosB C A +
=2 2
.... ∴ −
=
sin cos90
2 20 A A
Henceproved.
Introduction to Trigonometry
343
7. sin 670 + cos 750=cos(900 – 670)+sin(900 – 750) .... and
sin cos( º )cos sin( º )θ θθ θ= −= −
9090
= cos 230 + sin 150
Test Yourself – TRG 3
1. (i) 1 (ii) 1 (iii) 1 (iv) 0 (v) 2 (vi) 0
(vii) 1 (viii) 32
(ix) 0 (x) 1
(xi) 2
3. (i) sec12°+cot12° (ii) cosec6°+cos6°
(iii) sec34°+tan34° (iv) cosec5°+cos5°
(v) sin21°+tan21°
4. sec15°+tan15°
Exercise – 7.4 1. 1 + cot2 θ = cosec2 θ ⇒ cosec θ =± +1 2cot θ
⇒ cosec θ = 1 2+ cot θ
Now, sin θ = 1 11 2cosec θ θ
=+ cot
sincot
θθ
=+
11 2
sin2 θ + cos2 θ = 1
∴ cos2 θ = 1 – sin2 θ = 1 –1
1 2
2
+
cot θ
∴ cos2 θ = 1 – 11
1 11 12
2
2
2
2+=+ −+
=+cot
cotcot
cotcotθ
θθ
θθ
∴ cos cotcot
θ θ
θ=
+1 2
Introduction to Trigonometry
344
Now, sec θ = 1 1
1 2
cos cotcot
θ θ
θ
=
+
⇒ sec θ = 1 × 1 2+ cotcot
θ
θ
⇒ sec θ = 1 2+ cot
cotθ
θ
Also, tancot
θθ
= 1
2. 1 + tan2 A = sec2 A ∴ tan2 A = sec2 A – 1
∴ tan secA A= ± −2 1
∴ tan secA A== −−2 1
Now, cossec
AA
==1
tan sincos
A AA
=
∴ sin A = tan A × cos A
∴ sin A = secsec
2 1 1AA
− ×
sinsecsec
AAA
==−−2 1
cosec AA
=1
sin
∴ cosec AAA
=−
1
12secsec
cosec A AA
= ×−
112
secsec
∴ cosec AAA
=−
sec
sec2 1
cottan
AA
=1
Introduction to Trigonometry
345
∴ cotsec
AA
==−−
112
3. (i) sin sincos cos
cos ( º º ) sin ºsin (
2 0 2 0
2 0 2 0
2 2
2
63 2717 73
90 63 279
++
=− +
00 17 732 0º º ) cos− + ....
and sin cos( º )
cos sin( º )θ θθ θ= −= −
9090
=cos + sinsin 73 + cos 73
2 0 2 0
2 0 2 0
27 27 ....[ sin2 θ + cos2 θ=1]
=11
= 1 (ii) sin 250cos 650 + cos 250 sin 650
=cos(900 – 250).cos 650+sin(900 – 250) sin 650 .... cos = sin (90 )and sin = cos (90 )
θ θθ θ
−−
= cos 650.cos 650 + sin 650.sin 650 = cos2 650 + sin2 650
= 1 ....[ sin2 θ + cos2 θ=1] 4. (i) 9 sec2 A – 9 tan2 A = 9 (sec2 A – tan2 A) = 9 (1) ....
sec tan2 2 1A A− = = 9 ∴ Option (B) is correct
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
+ + + −1
11
1sincos cos
cossin sin
θθ θ
θθ θ
=+ +
+ −
cos sincos
sin cossin
θ θθ
θ θθ
1 1
=−(sin + cos ) 1
sin . cos
2θ θθ θ
= sin + cos + 2 sin . cos 1
sin . cos
2 2θ θ θ θθ θ
− ....[ sin2 θ + cos2 θ=1]
= 1 + 2 sin . cos 1
sin . cosθ θθ θ
−
=2 sin . cossin . cos
θ θθ θ
= 2 ∴ Option (C) is correct
Introduction to Trigonometry
346
(iii) (sec A + tan A) (1 – sin A) = +
−
1 1cos
sincos
( sin )A
AA
A
=+
−( sin )
cos( sin )1 1A
AA
=−1 2sincos
AA
....[1 – sin2 A = cos2 A]
=coscos
2 AA
=×cos cos
cosA A
A = cos A ∴ Option (D) is correct
(iv) 11
2
2
++
tancot
AA
=sec
cos
2
2
Aec A
.... 1
1+ =+ =
tan ² sec ²cot ² ²
cosec
A AA
= 1 12 2cos sinA A
√
= ×1
12
2
cossin
AA
=sincos
2
2
AA ....
sincos
tan2
22A
AA=
= tan2 A ∴ Option (D) is correct AlternateMethod:
11
2
2
++
tancot
AA
=
+
+
111
1
tan ²
tan ²
A
A
=
++
11
tan ²tan ²
tan ²
AA
A
= +( )×+( )
11
tan ² tan ²tan ²
A AA
= tan² A
Introduction to Trigonometry
347
5. (i) LHS = (cosec θ – cot θ)²
= 12
sincossinθ
θθ
−
=−
12
cossin
θθ
=−( cos )sin
1 2
2
θθ
= ( cos )cos
11
2
2
−−
θθ
=−−
−+
( cos )( cos )
( cos )( cos )
11
11
θθ
θθ
= 11−+
coscos
θθ
= RHS ∴ LHS = RHS Henceproved.
(ii) LHS =+
++cos
sinsin
cosAA
AA1
1
=+ +
+cos ( sin )
( sin ).cos
2 211
A AA A
=+ + ++
cos sin sin( sin ).cos
2 21 21A A A
A A
=+ + ++
(sin cos ) sin( sin ).cos
2 2 1 21
A A AA A
=+ ++
1 1 21
sin( sin ).cos
AA A
.... (∴ sin² A + cos² A = 1)
=+
+2 2
1sin
( sin ).cosA
A A
=+
+2 1
1( sin )
( sin ).cosA
A A
=
2 1
cos A
= 2 sec A = RHS ∴ LHS = RHS Henceproved.
Introduction to Trigonometry
348
(iii) LHS =−
+−
tancot
cottan
θθ
θθ1 1
= sin / cos
cossin
cos / sinsincos
θ θ
θθ
θ θθθ
( )
−
+−
11
11
=−
+−
sin / cossin cos
sin
cos / sincos sin
cos
θ θθ θ
θ
θ θθ θ
θ
= ×−
+ ×− −
sincos
sinsin cos
cossin
cos[ (sin cos )]
θθ
θθ θ
θθ
θθ θ
= sin ²
cos (sin cos )cos ²
sin (sin cos )
θ
θ θ θθ
θ θ θ−−
−
= sin ³ cos ³
cos .sin (sin cos )θ θ
θ θ θ θ−
−
= (sin cos ) (sin ² sin .cos cos ² )
cos .sin (sin cos )θ θ θ θ θ θ
θ θ θ θ
− + +
− ...[a³ – b³ = (a – b) (a² + ab + b²)]
=+1 sin .coscos .sin
θ θθ θ
...[sin²θ + cos² θ=1]
= +1
cos .sinsin .coscos .sinθ θ
θ θθ θ
= × +1 1 1
cos sinθ θ
= sec θ . cosec θ + 1 = RHS ∴ LHS = RHS Henceproved.
(iv) LHS = +1 secsec
AA
= +
÷
1 1 1
cos cosA A
=+
×1
1cos
coscosA
AA
= 1 + cos A
= ( cos ) ( cos )( cos )
1 11
+ ×−−
A AA ...[rationalisingthenumeratorby(1–cosA)]
Introduction to Trigonometry
349
=−−
11
2coscos
AA
= sin
cos
2
1AA−
= RHS ∴ LHS = RHS Henceproved. AlternateMethod:
LHS = +1 secsec
AA
= +
÷1 1 1
cos cosA A
=+
×
coscos
cosAA
A11
= cos A + 1
RHS =−sin
cos
2
1AA
=−−
11
2coscos
AA
=− +
−
( cos ) ( cos )( cos )
1 11
A AA
= 1 + cos A = cos A + 1 ∴ LHS = RHS Henceproved.
(v) LHS = cos sincos sin
A AA A− ++ −
11
DividingnumeratoranddenominatorbysinA
= cotcot
A AA A− ++ −
11
coseccosec
=+ − −
+ −(cot ) ( cot )
( cot )A A A A
A Acosec cosec
cosec
2 2
1 .... ( cosec² A – cot² A = 1)
=+ − + −
+ −(cot ) ( cot ) ( cot )
( cotA A A A A A
Acosec cosec cosec
cose1 cc A)
Introduction to Trigonometry
350
=+ − +
+ −(cot ) ( cot )
( cot )A A A A
A Acosec cosec
cosec1
1
= cot A + cosec A = RHS ∴ LHS = RHS Henceproved. AlternateMethod
LHS =− ++ −
cos sincos sin
A AA A
11
=− −+ −
cos (sin )cos (sin )
A AA A
11
=− −+ −
×− −− −
cos (sin )cos (sin )
cos (sin )cos (sin )
A AA A
A AA A
11
11 (byrationalizing)
=− −− −
[cos (sin )]cos (sin )
A AA A
11
2
2 2
=+ − − −
− + −cos (sin ) cos (sin )
cos (sin sin )
2 2
2 2
1 2 11 2
A A A AA A A
=+ + − − +
− − +cos sin sin cos sin cos
cos sin sin
2 2
2 2
1 2 2 21 2
A A A A A AA A A
=+ − + −
− − − +1 1 2 2 2
1 1 22 2
sin cos cos sinsin sin sin
A A A AA A A (∴ cos2A + sin2A = 1 and cos2A = 1 – sin2A)
= 2 2 2 1
2 22
− + −− +sin cos ( sin )
sin sinA A A
A A
= 2 1 1
2 1[( sin ) cos ( sin )]
sin ( sin )− + −
−A A A
A A
=− +
−
( sin ) ( cos )sin ( sin )
1 11
A AA A
=+
= +1 1cos
sin sincossin
AA A
AA
= cosec A + cot A
(vi) LHS =+−
11
sinsin
AA
=+−
×++
11
11
sinsin
sinsin
AA
AA (byrationalizing)
Introduction to Trigonometry
351
=+−
( sin )sin
11
2
2
AA
=+( sin )cos
1 2
2
AA
=+1 sincos
AA
=−−
sin 2sin2cos cos
3
3
θ θθ θ
= sec A + tan A = RHS ∴ LHS = RHS Henceproved.
(vii) LHS =−−
sin 2sin2cos cos
3
3
θ θθ θ
=−
−sin (1 2sin )cos (2cos 1)
2
2
θ θθ θ
=−− −
sin (1 2sin )cos 2(1 sin ) 1
2
2
θ θθ θ
=−
− −sin (1 2sin )
cos (2 2sin² 1)
2θ θθ θ
=−( )−( )
=sin sin ²
cos sin ²sincos
θ θ
θ θ
θθ
1 2
1 2
= tan θ = RHS ∴ LHS = RHS Henceproved. (viii) LHS = (sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + 2sin A cosec A + cosec2 A + cos2 A + 2cos A sec A + sec2 A = (sin2 A + cos2 A) + (cosec2 A) + (sec2 A) + 2sin A cosec A + 2 cos A sec A
= 1+ (1+ cot2 A) + (1 + tan2 A) + 2 sin A 1 2 1sin
coscos A
AA
+
....[ sin² A + cos² A = 1, cosec² A = 1 + cot² A, sec² A = 1 + tan² A] = 1 + 1 + cot² A + 1 + tan² A + 2 + 2 = 7 + tan2 A + cot2 A
Introduction to Trigonometry
352
= RHS ∴ LHS = RHS Henceproved. (ix) L.H.S = (cosec A – sin A) (sec A – cos A)
= −
−
1 1sin
sincos
cosA
AA
A
=
−
−
1 12 2sinsin
coscos
AA
AA
= ×
cossin
sincos
2 2AA
AA
= cos A . sin A ....(i)
RHS =+1
tan cotA A
=+
1sincos
cossin
AA
AA
=+−
12 2sin cos
cos sinA AA A
=+
cos .sinsin cos
A AA A2 2
= cos A . sin A ....(ii) ....(∴ sin2 A + cos2 A = 1) From (i) and (ii), LHS = RHS ∴ LHS = RHS Henceproved.
(x) Consider, 11
2
2 2
2++
=tancot
secAA
AAcosec
=
÷
1 12 2cos sinA A
= ×1
12
2
cossin
AA
∴ 11
2
2
++++
tancot
AA = tan² A ....(i)
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353
Consider, 11
1 12 2 2
−−
= −
÷ −
tancot
sincos
cossin
AA
AA
AA
=−
÷
−
cos sincos
sin cossin
A AA
A AA
2 2
= −−
× −
(sin cos )cos
sinsin cos
A AA
AA A
2 2
=−
×−
(sin cos )cos
sin(sin cos )
A AA
AA A
2
2
2
2
=sincos
2
2
AA
11
2−−−−
tancot
AA
= tan² A ....(ii)
From (i) and (ii),
11
11
2
2
22
++++
==−−−−
==
tancot
tancot
tanAA
AA
A