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If $\cosh \beta=\sec \alpha \cos \theta, \sinh \beta=\operatorname{cosec} \alpha \sin \theta$, then $\sinh ^2 \beta=$
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Verified Answer
The correct answer is:
$\sin ^2 \alpha$
We have,
$$
\begin{aligned}
& \cosh \beta=\sec \alpha \cos \theta \\
& \Rightarrow \quad \cos ^2 \mathrm{~h} \beta=\sec ^2 \alpha \cos ^2 \theta \\
& \Rightarrow \quad \cosh ^2 \beta=\sec ^2 \alpha\left(1-\sin ^2 \theta\right) \\
& \Rightarrow \quad \cosh ^2 \beta=\sec ^2 \alpha\left(1-\sinh ^2 \beta \sin ^2 \alpha\right) \\
& \Rightarrow \quad 1+\sinh ^2 \beta=\sec ^2 \alpha-\tan ^2 \alpha \sinh ^2 \beta \\
& \Rightarrow \quad \sinh ^2 \beta\left(1+\tan ^2 \alpha\right)=\sec ^2 \alpha-1 \\
& \Rightarrow \quad \sinh ^2 \beta=\frac{\tan ^2 \alpha}{1+\tan ^2 \alpha}=\sin ^2 \alpha \\
& {\left[\because 1+\tan ^2 \theta=\sec ^2 \theta\right]} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \cosh \beta=\sec \alpha \cos \theta \\
& \Rightarrow \quad \cos ^2 \mathrm{~h} \beta=\sec ^2 \alpha \cos ^2 \theta \\
& \Rightarrow \quad \cosh ^2 \beta=\sec ^2 \alpha\left(1-\sin ^2 \theta\right) \\
& \Rightarrow \quad \cosh ^2 \beta=\sec ^2 \alpha\left(1-\sinh ^2 \beta \sin ^2 \alpha\right) \\
& \Rightarrow \quad 1+\sinh ^2 \beta=\sec ^2 \alpha-\tan ^2 \alpha \sinh ^2 \beta \\
& \Rightarrow \quad \sinh ^2 \beta\left(1+\tan ^2 \alpha\right)=\sec ^2 \alpha-1 \\
& \Rightarrow \quad \sinh ^2 \beta=\frac{\tan ^2 \alpha}{1+\tan ^2 \alpha}=\sin ^2 \alpha \\
& {\left[\because 1+\tan ^2 \theta=\sec ^2 \theta\right]} \\
&
\end{aligned}
$$
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