# Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(C)

In this exercise, the problems are related to trigonometric ratios of complementary angles. Learning the right procedures to such problems enhances strong conceptual knowledge. Hence, the Selina Solutions for Class 10 Maths created by experts at BYJUâ€™S is the right resource for the purpose. The Concise Selina Solutions for Class 10 Maths Chapter 21 Trigonometrical Identities Exercise 21(C) solution PDF is given in the link below.

## Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(C) Download PDF

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