# ICSE Solutions for Selina Concise Chapter 21 Trigonometrical Identities Class 10 Maths

### Exercise 21(A)

Prove the following identities:

1. (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)

Solution

Hence, Proved

2. (1 + sin A)/(1 – sin A) = (cosec A + 1)/(cosec A – 1)

Solution

Hence, Proved

3. 1/(tan A + cot A) = cos A sin A

Solution

Taking L.H.S,

Hence, Proved

4. tan A – cot A = (1 – 2 cos2 A)/(sin A cos A)

Solution

Taking LHS,

Hence, Proved

5. sin4 A – cos4 A = 2 sin2 A – 1

Solution

Taking L.H.S,

sin4 A – cos4 A

= (sinA)2 – (cos2 A)2

= (sin2 A + cos2 A) (sin2 A – cos2 A)

= sin2A – cos2A

= sin2A – (1 – sin2A) [Since, cos2 A = 1 – sin2 A]

= 2sin2 A – 1

– Hence Proved

6. (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A

Solution

Taking L.H.S,

(1 – tan A)2 + (1 + tan A)2

= (1 + tan2 A + 2 tan A) + (1 + tan2 A – 2 tan A)

= 2 (1 + tan2 A)

= 2 sec2 A [Since, 1 + tan2 A = sec2 A]

Hence, Proved

7. cosec4 A – cosec2 A = cot4 A + cot2 A

Solution

cosec4 A – cosec2 A

= cosec2 A(cosec2 A – 1)

= (1 + cot2 A) (1 + cot2 A – 1)

= (1 + cot2 A) cot2 A

= cot4 A + cot2 A = R.H.S

Hence, Proved

8. sec A (1 – sin A) (sec A + tan A) = 1

Solution

Taking L.H.S,

sec A (1 – sin A) (sec A + tan A)

Hence, Proved

9. cosec A (1 + cos A) (cosec A – cot A) = 1

Solution

Taking L.H.S,

Hence, Proved

10. sec2 A + cosec2 A = sec2 A . cosec2 A

Solution

Taking L.H.S,

Hence, Proved

11. (1 + tan2 A) cot A/ cosecA = tan A

Solution

Taking L.H.S,

= RHS

Hence, Proved

12. tan2 A – sin2 A = tan2 A. sin2 A

Solution

Taking L.H.S,

tan2 A – sin2 A

Hence, Proved

13. cot2 A – cos2 A = cos2A. cot2A

Solution

Taking L.H.S,

cot2 A – cos2 A

Hence, Proved

14. (cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A

Solution

Taking L.H.S,

(cosec A + sin A) (cosec A – sin A)

= cosec2 A – sin2 A

= (1 + cotA) – (1 – cos2 A)

= cotA + cos2 A = R.H.S

Hence, Proved

15. (sec A – cos A)(sec A + cos A) = sin2 A + tan2 A

Solution

Taking L.H.S,

(sec A – cos A)(sec A + cos A)

= (sec2 A – cos2 A)

= (1 + tan2 A) – (1 – sin2 A)

= sin2 A + tan2 A = RHS

Hence, Proved

16. (cos A + sin A)2 + (cosA – sin A)2 = 2

Solution

Taking L.H.S,

(cos A + sin A)2 + (cosA – sin A)2

= cos2 A + sin2 A + 2cos A sin A + cos2 A – 2cosA.sinA

= 2 (cos2 A + sin2 A) = 2 = R.H.S

Hence, Proved

17. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1

Solution

Taking LHS,

(cosec A – sin A)(sec A – cos A)(tan A + cot A)

= RHS

Hence, Proved

18. 1/ sec A + tan A = sec A – tan A

Solution

Taking LHS,

= RHS

Hence, Proved

19. cosec A + cot A = 1/ cosec A – cot A

Solution

Taking LHS,

cosec A + cot A

= RHS

Hence, Proved

20. sec A – tan A/sec A + tan A = 1 – 2 secA tanA + 2 tan2 A

Solution

Taking LHS,

= 1 + tan2 A + tan2 A – 2 sec A tan A

= 1 – 2 sec A tan A + 2 tan2 A = RHS

Hence, Proved

21. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

Solution

Taking LHS,

(sin A + cosec A)2 + (cos A + sec A)2

= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2cos A sec A

= (sin2 A + cos2 A ) + cosec2 A + sec2 A + 2 + 2

= 1 + cosec2 A + sec2 A + 4

= 5 + (1 + cot2 A) + (1 + tan2 A)

= 7 + tan2 A + cot2 A = RHS

Hence, Proved

22. sec2 A. cosec2 A = tan2 A + cot2 A + 2

Solution

Taking,

RHS = tan2 A + cot2 A + 2 = tan2 A + cot2 A + 2 tan A. cot A

= (tan A + cot A)2 = (sin A/cos A + cos A/ sin A)2

= (sin2 A + cos2 A/ sin A.cos A)2 = 1/ cos2 A. sin2 A

= sec2 A. cosec2 A = LHS

Hence, Proved

23. 1/1 + cos A + 1/1 – cos A = 2 cosec2 A

Solution

Taking LHS,

= RHS

Hence, Proved

24. 1/1 – sin A + 1/1 + sin A = 2 sec2 A

Solution

Taking LHS,

= RHS

Hence, Proved

### Exercise 21(B)

1. Prove that:

(i) cos A/(1 – tan A) + sin A /(1 – cot A) = sin A + cos A

(ii) (cos3 A + sin3 A)/(cos A + sin A) + (cos3 A − sin3 A)/(cos A + sin A) = 2

(iii) tan A/(1 – cot A) + cot A/(1 –tan A) = sec A cosec A +1

(iv) (tan A + 1/cos A)2 + (tan A – 1/cos A)2 = 2 [(1+sin2 A)/(1 – sin2A)]

(v) 2 sin2 A + cos4 A = 1 + sin4 A

(vi) (sin A – sin B)/(cos A + cos B) + (cos A – cos B)/(sin A + sin B) = 0

(vii) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

(viii) (1+ tan A tan B)2 + (tan A – tan B)2 = sec2 A sec2 B

(ix) 1/(cos A + sin A -1) + 1/(cos A + sin A +1) = cosec A + sec A

Solution

(i)

Hence, Proved

(ii) Taking LHS,

Hence, Proved

(iii)

Hence, Proved

(iv)

Hence, Proved

(v) Taking LHS,

2 sin2 A + cos2 A

= 2 sin2 A + (1 – sin2 A)2

= 2 sin2 A+ 1 + sin4 A – 2 sin2 A

= 1 + sin4 A = RHS

Hence, Proved

(vi)

Hence, Proved

(vii)

Hence, Proved

(viii) (1 + tanA tanB)2 + (tanA – tanB)2

= 1 + tan2A tan2B + 2tanA tanB + tan2A + tan2B – 2tanA tanB

= 1 + tan2A + tan2B + tan2A tan2B

= sec2A + tan2B(1 + tan2A)

= sec2A + tan2B sec2A

= sec2A(1 + tan2B)

= sec2A sec2B

Hence, Proved

(ix)

Hence, Proved

2. If x cos A + y sin A = m and x sin A – y cos A = n, then prove that:

x2 + y2 = m2 + n2

Solution

Taking RHS,

m2 + n2

= (x cos A + y sin A)2 + (x sin A – y cos A)2

= x2 cos2 A + y2 sin2 A + 2xy cos A sin A + x2 sin2 A + y2 cos2 A – 2xy sin A cos A

= x2 (cos2 A + sin2 A) + y2 (sin2 A + cos2 A)

= x2 + y2 [Since, cos2 A + sin2 A = 1]

= RHS

3. If m = a sec A + b tan A and n = a tan A + b sec A, prove that m2 – n2 = a2 – b2

Solution

Taking LHS,

m2 – n2

= (a sec A + b tan A)2 – (a tan A + b sec A)2

= a2 sec2 A + b2 tan2 A + 2 ab sec A tan A – a2 tan2 A – b2 sec2 A – 2ab tan A sec A

= a2 (sec2 A – tan2 A) + b2 (tan2 A – sec2 A)

= a2 (1) + b2 (-1) [Since, sec2 A – tan2 A = 1]

= a2 – b2

= RHS

### Exercise 21(C)

1. Show that:

(i) tan 10o tan 15o tan 75o tan 80o = 1

Solution

Taking, tan 10o tan 15o tan 75o tan 80o

= tan (90o – 80o) tan (90o – 75o) tan 75o tan 80o

= cot 80o cot 75o tan 75o tan 80o

= 1 [Since, tan Î¸ × cot Î¸ = 1]

(ii) sin 42o sec 48o + cos 42o cosec 48o = 2

Solution

Taking, sin 42o sec 48o + cos 42o cosec 48o

= sin 42o sec (90o – 42o) + cos 42o cosec (90o – 42o)

= sin 42cosec 42+ cos 42o sec 42o

= 1 + 1 [Since, sin Î¸ × cosec Î¸ = 1 and cos Î¸ × sec Î¸ = 1]

= 2

(iii) sin 26o/sec 64+ cos 26o/cosec 64o = 1

Solution

Taking,

2. Express each of the following in terms of angles between 0°and 45°:

(i) sin 59°+ tan 63°

(ii) cosec 68°+ cot 72°

(iii) cos 74°+ sec 67°

Solution

(i) sin 59°+ tan 63°

= sin (90 – 31)°+ tan (90 – 27)°

= cos 31°+ cot 27°

(ii) cosec 68°+ cot 72°

= cosec (90 – 22)°+ cot (90 – 18)°

= sec 22°+ tan 18°

(iii) cos 74°+ sec 67°

= cos (90 – 16)°+ sec (90 – 23)°

= sin 16°+ cosec 23°

3. Show that:

(i) sin A/sin(90° −A) + cosA/cos(90° −A) = sec A cosec A

(ii) sin A cos A− sin Acos (90° −A)cosA/sec(90° −A) – cos Asin(90° −A)sin A/cosec (90° −A) = 0

Solution

= sin A cos A – sin3 A cos A – cos3 A sin A

= sin A cos A – sin A cos A (sin2 A + cos2 A)

= sin A cos A – sin A cos A (1) [Since, sin2 A + cos2 A = 1]

= 0

4. For triangle ABC, show that:

(i) sin (A + B)/2 = cos C/2

(ii) tan (B + C)/2 = cot A/2

Solution

We know that, in triangle ABC

∠A + ∠B + ∠C = 180o [Angle sum property of a triangle]

(i) Now,

(∠A + ∠B)/2 = 90o – ∠C/2

So, sin {(A + B)/2} = sin (90o – C/2)

= cos C/2

(ii) And,

(∠C + ∠B)/2 = 90o – ∠A/2

So,

tan ((B + C)/2) = tan (90o – A/2)

= cot A/2

5. Evaluate:

(i) 3(sin72°/cos 18°) – sec 32°/cosec 58°

(ii) 3cos80° cosec10° + 2 cos59°cosec 31°

(ii) sin80°/cos10° + sin59° sec 31°

(iv) tan(55° - A ) – cot(35° + A)

(v) cosec(65° + A) sec(25° A)

(vi) 2(tan57°/cot 33°) – cot70°/tan 20° 2 cos45°

(vii) cot241°/tan249° −2sin2 75°/cos2 15°

(viii) cos 70°/sin 20° +cos 59°/sin 31° − 8sin2 30°

(ix) 14 sin 30° + 6 cos 60° − 5tan45°

Solution

(i)

(ii) 3 cos 80o cosec 10o + 2 cos 59o cosec 31o

= 3 cos (90 – 10)cosec 10o + 2 cos (90 – 31)o cosec 31o

= 3 sin 10cosec 10o + 2 sin 31o cosec 31o

= 3 + 2 = 5

(iii) sin 80o/cos 10o + sin 59o sec 31o

= sin (90 – 10)o/cos 10o + sin (90 – 31)o sec 31o

= cos 10o/cos 10o + cos 31o sec 31o

= 1 + 1 = 2

(iv) tan (55o – A) – cot (35o + A)

= tan [90o – (35o + A)] – cot (35o + A)

= cot (35o + A)] – cot (35o + A)

= 0

(v) cosec (65o + A) – sec (25o – A)

= cosec [90o – (25o – A)] – sec (25o – A)

= sec (25o – A) – sec (25o – A)

= 0

(vi)

(vii)

= 1 – 2 = -1

(viii)

(ix) 14 sin 30o + 6 cos 60o – 5 tan 45o

= 14 (1/2) + 6 (1/2) – 5(1)

= 7 + 3 – 5

= 5

6. A triangle ABC is right angled at B; find the value of (sec A. cosec C – tan A. cot C)/ sin B

Solution

As, ABC is a right angled triangle right angled at B

So, A + C = 90o

(sec A. cosec C – tan A. cot C)/sin B

= (sec (90o – C). cosec C – tan (90o – C). cot C)/sin 90o

= (cosec C. cosec C – cot C. cot C)/1 = cosec2 C – cot2 C

= 1 [Since, cosec2 C – cot2 C = 1]

### Exercise 21(D)

1. Use tables to find sine of:

(i) 21°

(ii) 34° 42′

(iii) 47° 32′

(iv) 62° 57′

(v) 10° 20′ + 20° 45′

Solution

(i) sin 21o = 0.3584

(ii) sin 34o 42’= 0.5693

(iii) sin 47o 32’

= sin (47o 30′ + 2′)

=0.7373 + 0.0004

= 0.7377

(iv) sin 62o 57′

= sin (62o 54′ + 3′)

= 0.8902 + 0.0004

= 0.8906

(v) sin (10o 20′ + 2045′)

= sin 30o65′

= sin 31o5′

= 0.5150 + 0.0012

= 0.5162

2. Use tables to find cosine of:

(i) 2° 4’

(ii) 8° 12’

(iii) 26° 32’

(iv) 65° 41’

(v) 9° 23’ + 15° 54’

Solution

(i) cos 2° 4’ = 0.9994 – 0.0001 = 0.9993

(ii) cos 8° 12’ = cos 0.9898

(iii) cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 – 0.0003 = 0.8946

(iv) cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 -0.0013 = 0.4118

(v) cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 – 0.0006 = 0.9042

3. Use trigonometrical tables to find tangent of:

(i) 37°

(ii) 42° 18′

(iii) 17° 27′

Solution

(i) tan 37= 0.7536

(ii) tan 4218′ = 0.9099

(iii) tan 17o 27′ = tan (1724′ + 3′) = 0.3134 + 0.0010 = 0.3144

### Exercise 21(E)

1. Prove the following identities:

(i) 1/(cos A + sin A) + 1/(cos A – sin A) = 2cos A/(2cos2A – 1)

(ii) cosec A – cot A = sin A /(1+ cos A)

(iii) 1- sin2A/(1+ cos A) = cos A

(iv) (1- cos A)/sin A + sin A/ 1- cos A = 2 cosec A

(v) cot A /1 – tan A + tan A/ (1-cot A) = 1+ tan A + cot A

(vi) cos A/ (1+ sin A)+ tan A = sec A

(vii) sin A/(1-cos A) – cot A = cosec A

(viii) (sin A –cos A+1)/(sin A + cos A – 1) = cos A/(1-sin A)

(ix)√[(1+sin A)/(1 – sin A)] = cos A/ (1 – sinA)

(x) √[(1 – cos A)/(1+cos A) = sin A/(1 + cos A)

(xi) 1+(sec A – tan A)2 /cosec A(sec A – tan A) = 2 tan A

(xii) [(cosec A – cot A)2 + 1]/sec A(cosec A – cot A) = 2 cot A

(xiii) cot2 A{(sec A – 1 )/(1+ sin A)} + sec2 A{(sin A – 1 )/(1+ sec A)} = 0

(xiv) {(1- 2sin2 A)2/(cos 4A – sin4A) = 2 cos2 A – 1

(xv) sec4 A(1 –sin4 A)−2tan2A = 1

(xvi) cosec4A(1−cos4A)− 2cot2 A = 1

(xvii) (1+tan A + sec A)(1+ cotA – cosec A) = 2

Solution

(i) Taking LHS,

1/(cos A + sin A) + 1/(cos A – sin A)

= RHS

Hence, Proved

(ii) Taking LHS, cosec A – cot A

= RHS

Hence, Proved

(iii) Taking LHS, 1 – sin2 A/(1 + cos A)

= RHS

Hence. Proved

(iv) Taking LHS,

(1 – cos A)/sin A + sin A/(1 – cos A)

= RHS

Hence, Proved

(v) Taking LHS, cot A/(1 – tan A) + tan A/(1 – cot A)

= RHS

Hence, Proved

(vi) Taking LHS, cos A/ (1 + sin A) + tan A

= RHS

Hence, Proved

(vii) Consider LHS,

= (sin A/(1 – cos A)) – cot A

We know that, cot A = cos A/sin A

So,
= (sin2 A – cos A + cos2 A)/(1 – cos A) sin A

= (1–cos A)/(1–cos A) sin A

= 1/sin A

= cosec A

(viii) Taking LHS, (sin A – cos A + 1)/ (sin A + cos A – 1)

= RHS

Hence Proved

(ix) Taking LHS,

= RHS

Hence, Proved

(x) Taking LHS,

= RHS

Hence, Proved

(xi) Taking LHS,

= RHS

Hence, Proved

(xii) Taking LHS,

= RHS

Hence, Proved

(xiii) Taking LHS,

= RHS

Hence, Proved

(xiv) Taking LHS,

= RHS

Hence, Proved

(xv) Taking LHS,

sec4 A (1 – sin4 A) – 2 tan2 A

= sec4 A(1 – sin2 A) (1 + sin2 A) – 2 tan2 A

= sec4 A(cos2 A) (1 + sin2 A) – 2 tan2 A

= sec2 A + sin2 A/ cos2 A – 2 tan2 A

= sec2 A – tan2 A

= 1 = RHS

Hence, Proved

(xvi) cosec4 A(1 – cos4 A) – 2 cot2 A

= cosec4 A (1 – cos2 A) (1 + cos2 A) – 2 cot2 A

= cosec4 A (sin2 A) (1 + cos2 A) – 2 cot2 A

= cosec2 A (1 + cos2 A) – 2 cot2 A

= cosec2 A + cos2 A/sin2 A – 2 cot2 A

= cosec2 A + cot2 A – 2 cot2 A

= cosec2 A – cot2 A

= 1 = RHS

Hence, Proved

(xvii) (1 + tan A + sec A) (1 + cot A – cosec A)

= 1 + cot A – cosec A + tan A + 1 – sec A + sec A + cosec A – cosec A sec A

= 2 + cos A/sin A+ sin A/cos A – 1/(sin A cos A)

= 2 + (cos2 A + sin2 A)/sin A cos A – 1/(sin A cos A)

= 2 + 1/(sin A cos A) – 1/(sin A cos A)

= 2 = RHS

Hence, Proved

2. If sin A + cos A = p and sec A + cosec A = q, then prove that: q(p2 – 1) = 2p

Solution

Taking the LHS, we have

q(p2 – 1) = (sec A + cosec A) [(sin A + cos A)2 – 1]

= (sec A + cosec A) (sin2 A + cos2 A + 2 sin A cos A – 1)

= (sec A + cosec A) (1 + 2 sin A cos A – 1)

= (sec A + cosec A) (2 sin A cos A)

= 2sin A + 2 cos A

= 2p

3. If x = a cos Î¸ and y = b cot Î¸, show that: a2/x2 – b2/y2 = 1

Solution

Taking LHS,

a2/x2 – b2/y2

4. If sec A + tan A = p, show that: sin A = (p2 – 1)/ (p2 + 1)

Solution

Taking RHS, (p2 – 1)/(p2 + 1)

5. If tan A = n tan B and sin A = m sin B, prove that:

cos2 A = m2 – 1/n2 – 1

Solution

Given,

tan A = n tan B

n = tan A/ tan B

And, sin A = m sin B

m = sin A/ sin B

Now, taking RHS and substitute for m and n

m2 – 1/n2 – 1

6. (i) If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A

(ii) If 4 cos2 A – 3 = 0, show that: cos 3A = 4 cos2 A – 3 cos A

Solution

(i) Given, 2 sin A – 1 = 0

So, sin A = ½

We know, sin 30o = 1/2

Hence, A = 30o

Now, taking LHS

sin 3A = sin 3(30o) = sin 30o = 1

RHS = 3 sin 30o – 4 sin3 30o  = 3 (1/2) – 4 (1/2)3 = 3 – 4(1/8) = 3/2 – ½ = 1

∴ LHS = RHS

(ii) Given, 4 cos2 A – 3 = 0

4 cos2 A = 3

⇒ cos2 A = 3/4

⇒ cos A = √3/2

We know, cos 30o = √3/2

Hence, A = 30o

Now, taking

LHS = cos 3A = cos 3(30o) = cos 90o = 0

RHS = 4 cos3 A – 3 cos A = 4 cos3 30o – 3 cos 30o = 4 (√3/2)3 – 3 (√3/2)

= 4 (3√3/8) – 3√3/2

= 3√3/2 – 3√3/2

= 0

∴ LHS = RHS

7. Evaluate:

(i) 2(tan 35°/cot55°)2 + (cot 55°/tan 35°)2 –(sec 40°/cosec 50°)

(ii)sec 26° sin64° + cosec 33°/sec 57°

(iii) 5 sin66°/cos24° − 2cot 85°/tan5°

(iv) cos 40° cosec 50° + sin50° sec40°

(v) sin27° sin63° − cos63° cos27°

(vi) 3 sin72°/cos18° −sec32°/cosec58°

(vii) 3 cos 80° cosec10° + 2 cos59°cosec31°

(viii) cos75°/sin15° + sin12°/cos78° − cos18°/sin72°

Solution

(i)

= 2 (1)2 + 12 – 3

= 2 + 1 – 3 = 0

(ii)

= 1 + 1 = 2

(iii)

(iv) cos 40o cosec 50o + sin 50o sec 40o

= cos (90 – 50)o cosec 50o + sin (90 – 50)o sec 40o

= sin 50o cosec 50+ cos 40o sec 40o

= 1 + 1 = 2

(v) sin 27o sin 63o – cos 63o cos 27o

= sin (90 – 63)o sin 63o – cos 63o cos (90 – 63)o

= cos 63o sin 63o – cos 63o sin 63o

= 0

(vi)

(vii) 3 cos 80o cosec 10o + 2 cos 59cosec 31o

= 3 cos (90 – 10)o cosec 10o + 2 cos (90 – 31)cosec 31o

= 3 sin 10o cosec 10o + 2 sin 31cosec 31o

= 3 + 2 = 5

(viii)

8. Prove that:

(i) tan (55o + x) = cot (35o – x)

(ii) sec (70o – Î¸) = cosec (20o + Î¸)

(iii) sin (28o + A) = cos (62o – A)

(iv) 1/(1 + cos (90o – A)) + 1/(1 – cos (90o – A)) = 2 cosec2 (90o – A)

(v) 1/(1 + sin (90o – A)) + 1/(1 – sin (90o – A)) = 2 sec2 (90o – A)

Solution:

(i) tan (55o + x) = tan [90o – (35o – x)] = cot (35o – x)
(ii) sec (70o – Î¸) = sec [90o – (20o + Î¸)] = cosec (20o + Î¸)
(iii) sin (28o + A) = sin [90o – (62o – A)] = cos (62o – A)

(iv)

(v)