Question

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