# Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities

The science which deals with the measurements of triangles is called trigonometry. Trigonometrical ratios and its relations are used to prove trigonometric identities. Further, the trigonometrical ratios of complementary angles and the use of trigonometrical tables are the other topics covered in this chapter. As this chapter lays foundation to higher class Mathematics, students should acquire a strong grip over this chapter. For this purpose, BYJUâ€™S has created the Selina Solutions for Class 10 Mathematics prepared by expert faculty having vast academic experience. This also improves the problem-solving skills of students, which are crucial from an examination point of view. The Selina Solutions for Class 10 Mathematics Chapter 21 Trigonometrical Identities PDF are available exercise-wise in the links given below.

## Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Download PDF

### Exercises of Concise Selina Solutions Class 10 Maths Chapter 21 Trigonometrical Identities

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

## Access Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities

Exercise 21(A) Page No: 237

Prove the following identities:

1. sec A â€“ 1/ sec A + 1 = 1 â€“ cos A/ 1 + cos A

Solution:

– Hence Proved

2. 1 + sin A/ 1 â€“ sin A = cosec A + 1/ cosec A â€“ 1

Solution:

– Hence Proved

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

Solution:

Taking L.H.S,

– Hence Proved

4. tan A â€“ cot A = 1 â€“ 2 cos2 A/ sin A cos A

Solution:

Taking LHS,

– Hence Proved

5. sin4 A â€“ cos4 A = 2 sin2 A â€“ 1

Solution:

Taking L.H.S,

sin4 A â€“ cos4 A

= (sin2 A)2 â€“ (cos2 A)2

= (sin2 A + cos2 A) (sin2 A â€“ cos2 A)

= sin2A â€“ cos2A

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

= 2sin2 A â€“ 1

– Hence Proved

6. (1 â€“ tan A)2 + (1 + tan A)2 = 2 sec2 A

Solution:

Taking L.H.S,

(1 â€“ 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 â€“ cosec2 A = cot4 A + cot2 A

Solution:

cosec4 A â€“ cosec2 A

= cosec2 A(cosec2 A â€“ 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 â€“ sin A) (sec A + tan A) = 1

Solution:

Taking L.H.S,

sec A (1 â€“ sin A) (sec A + tan A)

– Hence Proved

9. cosec A (1 + cos A) (cosec A â€“ 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/ cosec2 A = tan A

Solution:

Taking L.H.S,

= RHS

– Hence Proved

12. tan2 A â€“ sin2 A = tan2 A. sin2 A

Solution:

Taking L.H.S,

tan2 A â€“ sin2 A

– Hence Proved

13. cot2 A â€“ cos2 A = cos2A. cot2A

Solution:

Taking L.H.S,

cot2 A â€“ cos2 A

– Hence Proved

14. (cosec A + sin A) (cosec A â€“ sin A) = cot2 A + cos2 A

Solution:

Taking L.H.S,

(cosec A + sin A) (cosec A â€“ sin A)

= cosec2 A â€“ sin2 A

= (1 + cot2 A) â€“ (1 â€“ cos2 A)

= cot2 A + cos2 A = R.H.S

– Hence Proved

15. (sec A â€“ cos A)(sec A + cos A) = sin2 A + tan2 A

Solution:

Taking L.H.S,

(sec A â€“ cos A)(sec A + cos A)

= (sec2 A â€“ cos2 A)

= (1 + tan2 A) â€“ (1 â€“ sin2 A)

= sin2 A + tan2 A = RHS

– Hence Proved

16. (cos A + sin A)2 + (cosA â€“ sin A)2 = 2

Solution:

Taking L.H.S,

(cos A + sin A)2 + (cosA â€“ sin A)2

= cos2 A + sin2 A + 2cos A sin A + cos2 A â€“ 2cosA.sinA

= 2 (cos2 A + sin2 A) = 2 = R.H.S

– Hence Proved

17. (cosec A â€“ sin A)(sec A â€“ cos A)(tan A + cot A) = 1

Solution:

Taking LHS,

(cosec A â€“ sin A)(sec A â€“ cos A)(tan A + cot A)

= RHS

– Hence Proved

18. 1/ sec A + tan A = sec A â€“ tan A

Solution:

Taking LHS,

= RHS

– Hence Proved

19. cosec A + cot A = 1/ cosec A â€“ cot A

Solution:

Taking LHS,

cosec A + cot A

= RHS

– Hence Proved

20. sec A â€“ tan A/ sec A + tan A = 1 â€“ 2 secA tanA + 2 tan2 A

Solution:

Taking LHS,

= 1 + tan2 A + tan2 A â€“ 2 sec A tan A

= 1 â€“ 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 â€“ cos A = 2 cosec2 A

Solution:

Taking LHS,

= RHS

– Hence Proved

24. 1/ 1 â€“ sin A + 1/ 1 + sin A = 2 sec2 A

Solution:

Taking LHS,

= RHS

– Hence Proved

Exercise 21(B) Page No: 237

1. Prove that:

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 â€“ sin2 A)2

= 2 sin2 A+ 1 + sin4 A â€“ 2 sin2 A

= 1 + sin4 A = RHS

– Hence Proved

(vi)

– Hence Proved

(vii)

– Hence Proved

(viii)

– Hence Proved

(ix)

– Hence Proved

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

x2 + y2 = m2 + n2Â

Solution:

Taking RHS,

m2 + n2

= (x cos A + y sin A)2 + (x sin A â€“ y cos A)2

= x2 cos2 A + y2 sin2 A + 2xy cos A sin A + x2 sin2 A + y2 cos2 A â€“ 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 â€“ n2 = a2 â€“ b2

Solution:

Taking LHS,

m2 â€“ n2

= (a sec A + b tan A)2 â€“ (a tan A + b sec A)2

= a2 sec2 A + b2 tan2 A + 2 ab sec A tan A â€“ a2 tan2 A â€“ b2 sec2 A â€“ 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 â€“ b2

= RHS

Exercise 21(C) Page No: 328

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¸ x 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 â€“ 42o) + cos 42o cosec (90o â€“ 42o)

= sin 42o cosec 42o + cos 42o sec 42o

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

= 2

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

Solution:

Taking,

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

(i) sin 59Â°+ tan 63Â°

(ii) cosec 68Â°+ cot 72Â°

(iii) cosÂ 74Â°+ sec 67Â°

Solution:

(i) sin 59Â°+ tan 63Â°

= sin (90 – 31)Â°+ tan (90 – 27)Â°

= cos 31Â°+ cot 27Â°

(ii) cosec 68Â°+ cot 72Â°

= cosec (90 – 22)Â°+ cot (90 – 18)Â°

= sec 22Â°+ tan 18Â°

(iii) cos 74Â°+ sec 67Â°

= cosÂ (90 – 16)Â°+ sec (90 – 23)Â°

= sinÂ 16Â°+ cosec 23Â°

3. Show that:

Solution:

= sin A cos A â€“ sin3 A cos A â€“ cos3 A sin A

= sin A cos A â€“ sin A cos A (sin2 A + cos2 A)

= sin A cos A â€“ 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

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

(i) Now,

(âˆ A + âˆ B)/ 2 = 90o – âˆ C/ 2

So,

sin ((A + B)/ 2) = sin (90o – C/ 2)

= cos C/ 2

(ii) And,

(âˆ C + âˆ B)/ 2 = 90o – âˆ A/ 2

So,

tan ((B + C)/ 2) = tan (90o – A/ 2)

= cot A/ 2

5. Evaluate:

Solution:

(i)

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

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

= 3 sin 10o cosec 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) â€“ cot (35o + A)

= tan [90o â€“ (35o + A)] â€“ cot (35o + A)

= cot (35o + A)] â€“ cot (35o + A)

= 0

(v) cosec (65o + A) â€“ sec (25o â€“ A)

= cosec [90o â€“ (25o â€“ A)] â€“ sec (25o â€“ A)

= sec (25o â€“ A) â€“ sec (25o â€“ A)

= 0

(vi)

(vii)

= 1 â€“ 2 = -1

(viii)

(ix) 14 sin 30o + 6 cos 60o â€“ 5 tan 45o

= 14 (1/2) + 6 (1/2) â€“ 5(1)

= 7 + 3 â€“ 5

= 5

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

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

= (cosec C. cosec C â€“ cot C. cot C)/ 1 = cosec2 C â€“ cot2 C

= 1 [Since, cosec2 C â€“ cot2 C = 1]

Exercise 21(D) Page No: 331

1. Use tables to find sine of:

(i) 21Â°

(ii) 34Â°Â 42′

(iii) 47Â°Â 32′

(iv) 62Â°Â 57′

(v) 10Â°Â 20′ + 20Â°Â 45′

Solution:

(i) sin 21oÂ = 0.3584

(ii) sin 34oÂ 42’= 0.5693

(iii) sin 47oÂ 32’= sin (47oÂ 30′ + 2′) =0.7373 + 0.0004 = 0.7377

(iv) sin 62oÂ 57′ = sin (62oÂ 54′ + 3′) = 0.8902 + 0.0004 = 0.8906

(v) sin (10oÂ 20′ + 20oÂ 45′) = sin 30o65′ = sin 31o5′ = 0.5150 + 0.0012 = 0.5162

2. Use tables to find cosine of:

(i) 2Â°Â 4â€™

(ii) 8Â°Â 12â€™

(iii) 26Â°Â 32â€™

(iv) 65Â°Â 41â€™

(v) 9Â°Â 23â€™ + 15Â°Â 54â€™

Solution:

(i) cos 2Â°Â 4â€™ = 0.9994 – 0.0001 = 0.9993

(ii) cos 8Â°Â 12â€™ = cos 0.9898

(iii) cos 26Â°Â 32â€™ = cos (26Â°Â 30â€™ + 2â€™) = 0.8949 – 0.0003 = 0.8946

(iv) cos 65Â°Â 41â€™ = cos (65Â°Â 36â€™ + 5â€™) = 0.4131 -0.0013 = 0.4118

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

3. Use trigonometrical tables to find tangent of:

(i) 37Â°

(ii) 42Â°Â 18′

(iii) 17Â°Â 27′

Solution:

(i) tan 37oÂ = 0.7536

(ii) tan 42oÂ 18′ = 0.9099

(iii) tan 17o Â 27′ = tan (17oÂ 24′ + 3′) = 0.3134 + 0.0010 = 0.3144

Exercise 21(E) Page No: 332

1. Prove the following identities:

Solution:

(i) Taking LHS,

1/ (cos A + sin A) + 1/ (cos A â€“ sin A)

= RHS

– Hence Proved

(ii) Taking LHS, cosec A â€“ cot A

= RHS

– Hence Proved

(iii) Taking LHS, 1 â€“ sin2 A/ (1 + cos A)

= RHS

– Hence Proved

(iv) Taking LHS,

(1 – cos A)/ sin A + sin A/ (1 â€“ cos A)

= RHS

– Hence Proved

(v) Taking LHS, cot A/ (1 â€“ tan A) + tan A/ (1 â€“ 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 â€“ sin4 A) â€“ 2 tan2 A

= sec4 A(1 â€“ sin2 A) (1 + sin2 A) â€“ 2 tan2 A

= sec4 A(cos2 A) (1 + sin2 A) â€“ 2 tan2 A

= sec2 A + sin2 A/ cos2 A â€“ 2 tan2 A

= sec2 A â€“ tan2 A

= 1 = RHS

– Hence Proved

(xvi) cosec4 A(1 â€“ cos4 A) â€“ 2 cot2 A

= cosec4 A (1 â€“ cos2 A) (1 + cos2 A) â€“ 2 cot2 A

= cosec4 A (sin2 A) (1 + cos2 A) â€“ 2 cot2 A

= cosec2 A (1 + cos2 A) â€“ 2 cot2 A

= cosec2 A + cos2 A/sin2 A â€“ 2 cot2 A

= cosec2 A + cot2 A â€“ 2 cot2 A

= cosec2 A â€“ cot2 A

= 1 = RHS

– Hence Proved

(xvii) (1 + tan A + sec A) (1 + cot A â€“ cosec A)

= 1 + cot A â€“ cosec A + tan A + 1 â€“ sec A + sec A + cosec A â€“ cosec A sec A

= 2 + cos A/sin A+ sin A/cos A â€“ 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 â€“ b2/ y2 = 1

Solution:

Taking LHS,

a2/ x2 â€“ 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 â€“ 1/ n2 â€“ 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 â€“ 1/ n2 â€“ 1

6. (i) If 2 sin A â€“ 1 = 0, show that:

sin 3A = 3 sin A â€“ 4 sin3 A

(ii) If 4 cos2 A â€“ 3 = 0, show that:

cos 3A = 4 cos2 A â€“ 3 cos A

Solution:

(i) Given, 2 sin A â€“ 1 = 0

So, sin A = Â½

We know, sin 30o = 1/2

Hence, A = 30o

Now, taking LHS

sin 3A = sin 3(30o) = sin 30o = 1

RHS = 3 sin 30o â€“ 4 sin3 30o = 3 (1/2) â€“ 4 (1/2)3 = 3 â€“ 4(1/8) = 3/2 â€“ Â½ = 1

Therefore, LHS = RHS

(ii) Given, 4 cos2 A â€“ 3 = 0

4 cos2 A = 3

cos2 A = 3/4

cos A = âˆš3/2

We know, cos 30o = âˆš3/2

Hence, A = 30o

Now, taking

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

RHS = 4 cos3 A â€“ 3 cos A = 4 cos3 30o â€“ 3 cos 30o = 4 (âˆš3/2)3 â€“ 3 (âˆš3/2)

= 4 (3âˆš3/8) – 3âˆš3/2

= 3âˆš3/2 – 3âˆš3/2

= 0

Therefore, LHS = RHS

7. Evaluate:

Solution:

(i)

= 2 (1)2 + 12 â€“ 3

= 2 + 1 â€“ 3 = 0

(ii)

= 1 + 1 = 2

(iii)

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

= cos (90 â€“ 50)o cosec 50o + sin (90 â€“ 50)o sec 40o

= sin 50o cosec 50o + cos 40o sec 40o

= 1 + 1 = 2

(v) sin 27o sin 63o â€“ cos 63o cos 27o

= sin (90 – 63)o sin 63o â€“ cos 63o cos (90 – 63)o

= cos 63o sin 63o â€“ cos 63o sin 63o

= 0

(vi)

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

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

= 3 sin 10o cosec 10o + 2 sin 31o cosec 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 â€“ A)

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

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

Solution:

(i) tan (55o + x) = tan [90o â€“ (35o – x)] = cot (35o – x)

(ii) sec (70oÎ¸) = sec [90o â€“ (20o + Î¸)] = cosec (20o + Î¸)

(iii) sin (28o + A) = sin [90o â€“ (62o – A)] = cos (62o – A)

(iv)

(v)

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2023-24 will be updated soon.