2....1. Express the trigonometric ratios sinθ , sec θ and tan θ in terms of cot θ. 2. Write all...

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.


318


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

+ −+ +


319


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.


320


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


321


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


322


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


323


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


324


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


327

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


328

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


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


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


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


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


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


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


335

(iii) FALSE,cosθisabbreviationusedforcosineθ. (iv) FALSE,cotismeaninglesswithoutanangleθ.

Here, cot side adjacent to

side opposite toθ

θθ

=∠∠

(v) FALSE,sinθisnevergreaterthan1.


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


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


=

−

− ( )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



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


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


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


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.


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


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.


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


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


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


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


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


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.


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)]


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)


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

=− −+ −

×− −− −



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)


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


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)


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

2....1. Express the trigonometric ratios sinθ , sec θ and tan θ in terms of cot θ. 2. Write all...

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Transcript of 2....1. Express the trigonometric ratios sinθ , sec θ and tan θ in terms of cot θ. 2. Write all...