Q65 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

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

  • A
    $\sin \alpha \cos \alpha$
  • B
    $\cos ^2 \alpha$
  • Correct answer
    C
    $\sin ^2 \alpha$
  • D
    $\sin \alpha+\cos \alpha$

Solution

We have, $$ \begin{aligned} & \cosh \beta=\sec \alpha \cos \theta \\ & \Rightarrow \quad \cos ^2 \mathrm{~h} \beta=\sec ^2 \alpha \cos ^2 \theta \\ & \Rightarrow \quad \cosh ^2 \beta=\sec ^2 \alpha\left(1-\sin ^2 \theta\right) \\ & \Rightarrow \quad \cosh ^2 \beta=\sec ^2 \alpha\left(1-\sinh ^2 \beta \sin ^2 \alpha\right) \\ & \Rightarrow \quad 1+\sinh ^2 \beta=\sec ^2 \alpha-\tan ^2 \alpha \sinh ^2 \beta \\ & \Rightarrow \quad \sinh ^2 \beta\left(1+\tan ^2 \alpha\right)=\sec ^2 \alpha-1 \\ & \Rightarrow \quad \sinh ^2 \beta=\frac{\tan ^2 \alpha}{1+\tan ^2 \alpha}=\sin ^2 \alpha \\ & {\left[\because 1+\tan ^2 \theta=\sec ^2 \theta\right]} \\ & \end{aligned} $$

Hence, the correct answer is option: 'C'.

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