If sin x cosh y = cos theta,cos x sinh y = si | vedclass

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(D) Given $	an x = \frac{3}{4}$.Since $	an^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{9}{16}$,we have $\sin^2 x = \frac{9}{25}$ and $\cos^2 x = \frac{1

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$\frac{16}{25}$

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(D) Given $	an x = \frac{3}{4}$.Since $	an^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{9}{16}$,we have $\sin^2 x = \frac{9}{25}$ and $\cos^2 x = \frac{16}{25}$.We are given $\sin x \cosh y = \cos 	heta$ and $\cos x \sinh y = \sin 	heta$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we substitute the given expressions:$(\sin x \cosh y)^2 + (\cos x \sinh y)^2 = 1$$\sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y = 1$Using $\cosh^2 y = 1 + \sinh^2 y$,we get:$\sin^2 x (1 + \sinh^2 y) + \cos^2 x \sinh^2 y = 1$$\frac{9}{25}(1 + \sinh^2 y) + \frac{16}{25} \sinh^2 y = 1$Multiply by $25$:$9 + 9 \sinh^2 y + 16 \sinh^2 y = 25$$25 \sinh^2 y = 25 - 9$$25 \sinh^2 y = 16$$\sinh^2 y = \frac{16}{25}$.

If $\sin x \cosh y = \cos \theta$,$\cos x \sinh y = \sin \theta$ and $4 \tan x = 3$,then $\sinh^2 y =$

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