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

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Aicharya · Accepted Answer

$\begin{aligned} & \text {Since, } \tan x=\frac{3}{4} \ & \Rightarrow \quad \sin ^2 x=\frac{9}{25} \ & \text { and } \quad \cos ^2 x=\frac{16}{25} \ & \text { and } \sin x \cosh y=\cos \theta, \cos x \sinh y=\sin \theta \ & \because \cos ^2 \theta+\sin ^2 \theta=1 \ & \Rightarrow(\sin x \cosh y)^2+(\cos x \sinh y)^2=1 \ & \Rightarrow \frac{9}{25}\left(1+\sinh ^2 y\right)+\frac{16}{25} \sinh ^2 y=1 \ & \Rightarrow 9+9 \sinh ^2 y+16 \sinh ^2 y=25 \end{aligned}$ $\Rightarrow \sinh ^2 y=\frac{16}{25}$ Hence, option (4) is correct.

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

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