If sin x cosh y = cos theta and cos x sinh y | vedclass

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(B) Given equations are: $(1)$ $\sin x \cosh y = \cos 	heta$ $(2)$ $\cos x \sinh y = \sin 	heta$ Squaring both equations: $(1)$ $\sin^2 x \cosh^2 y

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$\cos^2 x$

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(B) Given equations are: $(1)$ $\sin x \cosh y = \cos 	heta$ $(2)$ $\cos x \sinh y = \sin 	heta$ Squaring both equations: $(1)$ $\sin^2 x \cosh^2 y = \cos^2 	heta$ $(2)$ $\cos^2 x \sinh^2 y = \sin^2 	heta$ We know that $\cosh^2 y - \sinh^2 y = 1$,so $\cosh^2 y = 1 + \sinh^2 y$. Substitute this into the first squared equation: $\sin^2 x (1 + \sinh^2 y) = \cos^2 	heta$ $\sin^2 x + \sin^2 x \sinh^2 y = \cos^2 	heta$ From $(2)$,$\sin^2 	heta = \cos^2 x \sinh^2 y$. Using $\cos^2 	heta = 1 - \sin^2 	heta$: $\sin^2 x + \sin^2 x \sinh^2 y = 1 - \cos^2 x \sinh^2 y$ $\sinh^2 y (\sin^2 x + \cos^2 x) = 1 - \sin^2 x$ Since $\sin^2 x + \cos^2 x = 1$ and $1 - \sin^2 x = \cos^2 x$: $\sinh^2 y = \cos^2 x$.

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

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