Question

If $\sin x \cos h y = \cos \theta , \cos x \sin hy = \sin \theta $ and $ 4 \tan x = 3$. Then $\sin h^2 y = $

• A. $\frac{4}{5}$
• B. $\frac{9}{16}$
• C. $\frac{9}{25}$
• D. $\frac{16}{25}$

Answer 1

Answer: (D)

Since, $\tan x=\frac{3}{4}$
$\Rightarrow \sin ^{2} x=\frac{9}{25}$
and $\cos ^{2} x=\frac{16}{25}$
and $\sin x \cosh y=\cos \theta, \cos x \sinh y=\sin \theta$
$\because \cos ^{2} \theta+\sin ^{2} \theta=1$
$\Rightarrow (\sin x \cosh y)^{2}+(\cos x \sin h y)^{2}=1$
$\Rightarrow \frac{9}{25}\left(1+\sinh ^{2} y\right)+\frac{16}{25} \sin h^{2} y=1$
$\Rightarrow 9+9 \sinh ^{2} y+16 \sinh ^{2} y=25$
$\Rightarrow \sinh ^{2} y=\frac{16}{25}$