Q146 / 200 · TS EAMCET 2001 · Mathematics

$\sin \mathrm{h}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is equal to

  • A
    $\operatorname{coth}^{-1} x$
  • B
    $\sin h^{-1} x$
  • C
    $-\tanh ^{-1} x$
  • Correct answer
    D
    $\tan \mathrm{h}^{-1} x$

Solution

We have, $\begin{aligned} & \quad \sin \mathrm{h}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=\theta \\ \Rightarrow \quad & \frac{x}{\sqrt{1-x^2}}=\sinh \theta\end{aligned}$ $\begin{aligned} & \text { Now, } \cosh ^2 \theta=1+\sin ^2 \theta \\ & =1+\frac{x^2}{1-x^2}=\frac{1}{1-x^2} \\ & \therefore \quad \cosh \theta=\frac{1}{\sqrt{1-x^2}} \\ & \Rightarrow \quad \tan \mathrm{h} \theta=\frac{\sinh \mathrm{h}}{\cosh \theta}=x \\ & \theta=\tan \mathrm{h}^{-1} x \\ & \Rightarrow \quad \sin \mathrm{h}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=\tanh ^{-1} x \\ & \end{aligned}$

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

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