If sin x cdot cosh y = cos theta and cos x cd | vedclass

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(B) Given: $\sin x \cdot \cosh y = \cos 	heta$  $(i)$ $\cos x \cdot \sinh y = \sin 	heta$  $(ii)$ Squaring and adding $(i)$ and $(ii)$: $(\sin x \cd

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$2$

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(B) Given: $\sin x \cdot \cosh y = \cos 	heta$  $(i)$ $\cos x \cdot \sinh y = \sin 	heta$  $(ii)$ Squaring and adding $(i)$ and $(ii)$: $(\sin x \cdot \cosh y)^2 + (\cos x \cdot \sinh y)^2 = \cos^2 	heta + \sin^2 	heta$ $\sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y = 1$ Using $\cosh^2 y = 1 + \sinh^2 y$ and $\cos^2 x = 1 - \sin^2 x$: $\sin^2 x (1 + \sinh^2 y) + (1 - \sin^2 x) \sinh^2 y = 1$ $\sin^2 x + \sin^2 x \sinh^2 y + \sinh^2 y - \sin^2 x \sinh^2 y = 1$ $\sin^2 x + \sinh^2 y = 1$ Since $\cosh^2 y = 1 + \sinh^2 y$,we have $\sinh^2 y = \cosh^2 y - 1$. Substituting this into the equation: $\sin^2 x + (\cosh^2 y - 1) = 1$ $\sin^2 x + \cosh^2 y = 2$

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

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