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 (2θ + 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.