# Class 11 RD Sharma Solutions – Chapter 8 Transformation Formulae – Exercise 8.2 | Set 2

**Question 11. If cosec A + sec A = cosec B + sec B, prove that tan A tan B =****.**

**Solution:**

We have, cosec A + sec A = cosec B + sec B

=> sec A − sec B = cosec B − cosec A

=>

=>

=>

=> tan A tan B =

=> tan A tan B =

Hence proved.

**Question 12. If sin 2A = λ sin 2B, prove that****.**

**Solution:**

We are given, sin 2A = λ sin 2B

=>

On applying Componendo and Dividendo, we get,

=>

=>

=>

=>

=>

Hence proved.

**Question 13. Prove that:**

**(i) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

=

= cot C

= R.H.S.

Hence proved.

**(ii) sin (B−C) cos (A−D) + sin (C−A) cos (B−D) + sin (A−B) cos (C−D) = 0**

**Solution:**

We have, L.H.S. = sin (B−C) cos (A−D) + sin (C−A) cos (B−D) + sin (A−B) cos (C−D)

=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 14. If****, prove that tan A tan B tan C tan D = −1.**

**Solution:**

We have,

=>

=>

=>

=>

=>

=>

=>

=>. . . . (1)

Also,

=>

=>

=>

=>

=>. . . . (2)

Dividing (1) by (2), we get,

=>

=>

=>

=> tan A tan B tan C tan D = −1

Hence proved.

**Question 15. If cos (**α+β) sin(γ+δ) = cos (α−β) sin(γ−δ), prove that cot α cot β cot γ = cot δ.

**Solution:**

We have, cos (α+β) sin(γ+δ) = cos (α−β) sin(γ−δ)

=>

=>

=>

=>

=>. . . . (1)

Also,

=>

=>

=>

=>. . . . (2)

Dividing (1) by (2), we get,

=>

=>

=> cot α cot β = tan γ cot δ

=> cot α cot β cot γ = cot δ

Hence proved.

**Question 16. If y sin Ø = x sin (2**θ** + Ø), prove that (x + y) cot (**θ** + Ø) = (y − x) cot **θ.

**Solution:**

Given, y sin Ø = x sin (2θ + Ø)

=>

On applying Componendo and Dividendo, we get,

=>

=>

=>

=>

=> tan (Ø+θ) cot θ =

=>

=> (y − x) cot θ = (x + y) cot (θ + Ø)

=> (x + y) cot (θ + Ø) = (y − x) cot θ

Hence proved.

**Question 17. If cos (A+B) sin (C**−**D) = cos (A−B) sin (C+D), prove that tan A tan B tan C + tan D = 0.**

**Solution:**

We are given, cos (A+B) sin (C−D) = cos (A−B) sin (C+D)

=>

On applying Componendo and Dividendo, we get,

=>

=>

=>

=>

=> −tan D = tan A tan B tan C

=> tan A tan B tan C + tan D = 0

Hence proved.

**Question 18. If ****, prove that xy + yz + zx = 0.**

**Solution:**

We have,= k (say)

x =

y =

z =

So, L.H.S. = xy + yz + zx

=

=

=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 19. If m sin **θ** = n sin (**θ** + 2a), prove that****.**

**Solution:**

We are given, m sin θ = n sin (θ + 2a)

=>

On applying Componendo and Dividendo, we get,

=>

=>

=>

=>

=>

Hence, proved.

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