Question

The value of ${\sinh }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is

• A. $\tan {\mathrm{h}}^{-1}\left(x\right)$
• B. $\cot {\mathrm{h}}^{-1}\left(x\right)$
• C. $\sin {\mathrm{h}}^{-1}\left(2x\right)$
• D. $\cos {\mathrm{h}}^{-1}\left(2x\right)$

Answer 1

The correct option is A

$\tan {\mathrm{h}}^{-1}\left(x\right)$

Explanation for the correct option

The given trigonometric expression: ${\sinh }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$.

Let us assume that, ${\sinh }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=y$

$\Rightarrow \sinh \left(y\right)=\frac{x}{\sqrt{1-x^2}}$

It is known that, $\mathbf{cos}{\mathbf{h}}^{\mathbf{2}}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathbf{-}\mathbf{sin}{\mathbf{h}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{1}$.

Thus, ${\cosh }^2\left(y\right)-\sin h^2\left(y\right)=1$.

$$\begin{gathered} \Rightarrow {\cosh }^2\left(y\right)-{\left(\frac{x}{\sqrt{1-x^2}}\right)}^2=1 \\ \Rightarrow {\cosh }^2\left(y\right)-\frac{x^2}{1-x^2}=1 \\ \Rightarrow {\cosh }^2\left(y\right)=1+\frac{x^2}{1-x^2} \\ \Rightarrow {\cosh }^2\left(y\right)=\frac{1-x^2+x^2}{1-x^2} \\ \Rightarrow {\cosh }^2\left(y\right)=\frac{1}{1-x^2} \\ \Rightarrow \cosh \left(y\right)=\frac{1}{\sqrt{1-x^2}} \end{gathered}$$

Since, $\tanh \left(y\right)=\frac{\sinh \left(y\right)}{\cos h\left(y\right)}$

$$\begin{gathered} \Rightarrow \tanh \left(y\right)=\frac{{\displaystyle \frac{x}{\sqrt{1-x^2}}}}{{\displaystyle \frac{1}{\sqrt{1-x^2}}}} \\ \Rightarrow \tanh \left(y\right)=x \\ \Rightarrow y={\tanh }^{-1}\left(x\right) \\ \Rightarrow {\sinh }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)={\tanh }^{-1}\left(x\right)\left[\because {\sinh }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=y\right] \end{gathered}$$

Therefore, the value of ${\sinh }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is $\tan {\mathrm{h}}^{-1}\left(x\right)$.

Hence, the correct option is (A).