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

 

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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:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 1

– Hence Proved

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

Solution:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 2

– Hence Proved

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

Solution:

Taking L.H.S,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 3

– Hence Proved

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

Solution:

Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 4

– 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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 5

– Hence Proved

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

Solution:

Taking L.H.S,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 6

– Hence Proved

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

Solution:

Taking L.H.S,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 7

– Hence Proved

11. (1 + tan2 A) cot A/ cosec2 A = tan A

Solution:

Taking L.H.S,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 8

= RHS

– Hence Proved

12. tan2 A – sin2 A = tan2 A. sin2 A

Solution:

Taking L.H.S,

tan2 A – sin2 A

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 9

– Hence Proved

13. cot2 A – cos2 A = cos2A. cot2A

Solution:

Taking L.H.S,

cot2 A – cos2 A

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 10

– 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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 11

= RHS

– Hence Proved

18. 1/ sec A + tan A = sec A – tan A

Solution:

Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 12

= RHS

– Hence Proved

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

Solution:

Taking LHS,

cosec A + cot A

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 13

= RHS

– Hence Proved

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

Solution:

Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 14

= 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,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 15

= RHS

– Hence Proved

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

Solution:

Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(A) - 16

= RHS

– Hence Proved


Exercise 21(B) Page No: 237

1. Prove that:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 1

Solution:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 2(i)

– Hence Proved

(ii) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 3

– Hence Proved

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 4(iii)

– Hence Proved

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 5(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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 6(vi)

– Hence Proved

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 7(vii)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 8

– Hence Proved

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 9(viii)

– Hence Proved

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(B) - 10(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 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 θ x cot θ = 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 42o cosec 42o + cos 42o sec 42o

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

= 2

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

Solution:

Taking,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 1

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:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 2

Solution:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 3

= 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:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 4

Solution:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 5(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) – 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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 6(vi)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 7(vii)

= 1 – 2 = -1

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(C) - 8(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) 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 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) Page No: 332

1. Prove the following identities:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 1

Solution:

(i) Taking LHS,

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 2

= RHS

– Hence Proved

(ii) Taking LHS, cosec A – cot A

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 3

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 4

= RHS

– Hence Proved

(iv) Taking LHS,

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 5

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 6

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 7

= RHS

– Hence Proved

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 8

= RHS

– Hence Proved

(ix) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 9

= RHS

– Hence Proved

(x) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 10

= RHS

– Hence Proved

(xi) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 11

= RHS

– Hence Proved

(xii) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 12

= RHS

– Hence Proved

(xiii) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 13

= RHS

– Hence Proved

(xiv) Taking LHS,

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 14

= 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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 15

4. If sec A + tan A = p, show that:

sin A = (p2 – 1)/ (p2 + 1)

Solution:

Taking RHS, (p2 – 1)/ (p2 + 1)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 25

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

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 16

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

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 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

Therefore, LHS = RHS

7. Evaluate:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 17

Solution:

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 18(i)

= 2 (1)2 + 12 – 3

= 2 + 1 – 3 = 0

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 19(ii)

= 1 + 1 = 2

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 20(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 50o + 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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 21

(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)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 22

8. Prove that:

(i) tan (55o + x) = cot (35o – x)

(ii) sec (70o – θ) = cosec (20o + θ)

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

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

(iv)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 23

(v)

Selina Solutions Concise Class 10 Maths Chapter 21 ex. 21(E) - 24

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