If cosh beta = sec alpha cos theta and sinh b | vedclass

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(C) Given: $\cosh \beta = \sec \alpha \cos 	heta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin 	heta$. From the second equation,$\sin 	heta

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$\sin^2 \alpha$

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(C) Given: $\cosh \beta = \sec \alpha \cos 	heta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin 	heta$. From the second equation,$\sin 	heta = \sinh \beta \sin \alpha$. Squaring both sides,$\sin^2 	heta = \sinh^2 \beta \sin^2 \alpha$. We know that $\cosh^2 \beta - \sinh^2 \beta = 1$,so $\cosh^2 \beta = 1 + \sinh^2 \beta$. Substitute $\cosh \beta = \sec \alpha \cos 	heta$ into the identity: $\sec^2 \alpha \cos^2 	heta = 1 + \sinh^2 \beta$. $\sec^2 \alpha (1 - \sin^2 	heta) = 1 + \sinh^2 \beta$. Substitute $\sin^2 	heta = \sinh^2 \beta \sin^2 \alpha$: $\sec^2 \alpha (1 - \sinh^2 \beta \sin^2 \alpha) = 1 + \sinh^2 \beta$. $\sec^2 \alpha - \sec^2 \alpha \sin^2 \alpha \sinh^2 \beta = 1 + \sinh^2 \beta$. Since $\sec^2 \alpha \sin^2 \alpha = 	an^2 \alpha$: $\sec^2 \alpha - 	an^2 \alpha \sinh^2 \beta = 1 + \sinh^2 \beta$. $\sec^2 \alpha - 1 = \sinh^2 \beta (1 + 	an^2 \alpha)$. $	an^2 \alpha = \sinh^2 \beta \sec^2 \alpha$. $\sinh^2 \beta = \frac{	an^2 \alpha}{\sec^2 \alpha} = \sin^2 \alpha$.

If $\cosh \beta = \sec \alpha \cos \theta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin \theta$,then $\sinh^2 \beta =$

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