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, BYJU’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 – 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

= (sin² A)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/ cosec² A = 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 + cot² A) – (1 – cos² A)

= cot² A + 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) 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 – 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)

– 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) Page No: 328

1. Show that:

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

Solution:

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

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

= cot 80^o cot 75^o tan 75^o tan 80^o

= 1 [Since, tan θ x cot θ = 1]

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

Solution:

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

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

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

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

= 2

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

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 = 180^o [Angle sum property of a triangle]

(i) Now,

(∠A + ∠B)/ 2 = 90^o – ∠C/ 2

So,

sin ((A + B)/ 2) = sin (90^o – C/ 2)

= cos C/ 2

(ii) And,

(∠C + ∠B)/ 2 = 90^o – ∠A/ 2

So,

tan ((B + C)/ 2) = tan (90^o – A/ 2)

= cot A/ 2

5. Evaluate:

Solution:

(i)

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

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

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

= 3 + 2 = 5

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

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

= cos 10^o/ cos 10^o + cos 31^o sec 31^o

= 1 + 1 = 2

(iv) tan (55^o – A) – cot (35^o + A)

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

= cot (35^o + A)] – cot (35^o + A)

= 0

(v) cosec (65^o + A) – sec (25^o – A)

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

= sec (25^o – A) – sec (25^o – A)

= 0

(vi)

(vii)

= 1 – 2 = -1

(viii)

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

= 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 = 90^o

(sec A. cosec C – tan A. cot C)/ sin B

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

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

= 1 [Since, cosec² C – cot² C = 1]

Exercise 21(D) Page No: 331

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 21^o = 0.3584

(ii) sin 34^o 42’= 0.5693

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

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

(v) sin (10^o 20′ + 20^o 45′) = sin 30^o65′ = sin 31^o5′ = 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^o = 0.7536

(ii) tan 42^o 18′ = 0.9099

(iii) tan 17^o  27′ = tan (17^o 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 – 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 θ and y = b cot θ, 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 30^o = 1/2

Hence, A = 30^o

Now, taking LHS

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

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

Therefore, LHS = RHS

(ii) Given, 4 cos² A – 3 = 0

4 cos² A = 3

cos² A = 3/4

cos A = √3/2

We know, cos 30^o = √3/2

Hence, A = 30^o

Now, taking

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

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

= 4 (3√3/8) – 3√3/2

= 3√3/2 – 3√3/2

= 0

Therefore, LHS = RHS

7. Evaluate:

Solution:

(i)

= 2 (1)² + 1² – 3

= 2 + 1 – 3 = 0

(ii)

= 1 + 1 = 2

(iii)

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

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

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

= 1 + 1 = 2

(v) sin 27^o sin 63^o – cos 63^o cos 27^o

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

= cos 63^o sin 63^o – cos 63^o sin 63^o

= 0

(vi)

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

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

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

= 3 + 2 = 5

(viii)

8. Prove that:

(i) tan (55^o + x) = cot (35^o – x)

(ii) sec (70^o – θ) = cosec (20^o + θ)

(iii) sin (28^o + A) = cos (62^o – A)

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

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

Solution:

(i) tan (55^o + x) = tan [90^o – (35^o – x)] = cot (35^o – x)

(ii) sec (70^o – θ) = sec [90^o – (20^o + θ)] = cosec (20^o + θ)

(iii) sin (28^o + A) = sin [90^o – (62^o – A)] = cos (62^o – 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.