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 BYJU’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.

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

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]