191.

If $2{\sinh }^{-1}\left(\frac{\mathrm{a}}{\sqrt{1 - {\mathrm{a}}^2}}\right) = \log \left(\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right)$, then x is equal to

• a

• $\frac{1}{\mathrm{a}}$

• $\sqrt{1 - {\mathrm{a}}^2}$

• $\frac{1}{\sqrt{1 - {\mathrm{a}}^2}}$

$$\begin{gathered} \mathrm{Given} \mathrm{that}, \\ 2{\sinh }^{-1}\left(\frac{\mathrm{a}}{\sqrt{1 - {\mathrm{a}}^2}}\right) = \log \left(\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right) \\ \Rightarrow 2{\sinh }^{-1}\left(\mathrm{x}\right) = 2\log \left(\mathrm{x} + \sqrt{1 + {\mathrm{x}}^2}\right) \\ \left[\because {\sinh }^{-1}\left(\mathrm{x}\right) = \log \left(\mathrm{x} + \sqrt{1 + {\mathrm{x}}^2}\right)\right] \\ = \log {\left(\mathrm{x} + \sqrt{1 + {\mathrm{x}}^2}\right)}^2 \\ = \log \left({\mathrm{x}}^2 + 1 + {\mathrm{x}}^2 + 2\mathrm{x}\sqrt{1 + {\mathrm{x}}^2}\right) \\ 2{\sinh }^{-1}\left(\mathrm{x}\right) = \log \left(2{\mathrm{x}}^2 + 1 + 2\mathrm{x}\sqrt{1 + {\mathrm{x}}^2}\right) \\ = \log \left[\frac{2\mathrm{a}}{1 - {\mathrm{a}}^2} + 1 + 2 \times \frac{\mathrm{a}}{\sqrt{1 - {\mathrm{a}}^2}} \times \sqrt{1 + \frac{{\mathrm{a}}^2}{1 - {\mathrm{a}}^2}}\right] \\ \left[\because \mathrm{x} = \frac{\mathrm{a}}{\sqrt{1 - {\mathrm{a}}^2}}\right] \\ = \log \left[\frac{2\mathrm{a}}{1 - {\mathrm{a}}^2} + 1 + \frac{2\mathrm{a}}{\sqrt{1 - {\mathrm{a}}^2}} \times \frac{1}{\sqrt{1 - {\mathrm{a}}^2}} \right] \\ = \log \left[\frac{2\mathrm{a}}{1 - {\mathrm{a}}^2} + 1 + \frac{2\mathrm{a}}{1 - {\mathrm{a}}^2}\right] \\ = \log \left[\frac{2{\mathrm{a}}^2 + 1 - {\mathrm{a}}^2 + 2\mathrm{a}}{1 - {\mathrm{a}}^2}\right] \\ = \log \left[\frac{{\mathrm{a}}^2 + 2\mathrm{a} + 1}{1 - {\mathrm{a}}^2}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{Put} \mathrm{the} \mathrm{value} \mathrm{of} \\ 2{\sinh }^{-1}\left(\frac{\mathrm{a}}{\sqrt{1 - {\mathrm{a}}^2}}\right), \mathrm{we} \mathrm{get} \\ \log \left[\frac{{\mathrm{a}}^2 + 2\mathrm{a} + 1}{1 - {\mathrm{a}}^2}\right] = \log \left[\frac{1 + \mathrm{x}}{1 - \mathrm{x}}\right] \\ \Rightarrow \frac{{\mathrm{a}}^2 + 2\mathrm{a} + 1}{1 - {\mathrm{a}}^2} = \frac{1 + \mathrm{x}}{1 - \mathrm{x}} \\ \Rightarrow \frac{{\mathrm{a}}^2 + 2\mathrm{a} + 1 + 1 - {\mathrm{a}}^2}{{\mathrm{a}}^2 + 2\mathrm{a} + 1 - 1 + {\mathrm{a}}^2} = \frac{1 + \mathrm{x} + 1 - \mathrm{x}}{1 - \mathrm{x} - 1 + \mathrm{x}} \\ \Rightarrow \frac{2 + 2\mathrm{a}}{2{\mathrm{a}}^2 + 2\mathrm{a}} = \frac{1}{\mathrm{x}} \\ \left[\because \mathrm{by} \mathrm{componendo} \mathrm{and} \mathrm{divdendo} \mathrm{rule}\right] \\ \Rightarrow \frac{2\left(1 + \mathrm{a}\right)}{2\mathrm{a}\left(1 + \mathrm{a}\right)} = \frac{1}{\mathrm{x}} \\ \Rightarrow \frac{1}{\mathrm{a}} = \frac{1}{\mathrm{x}} \\ \Rightarrow \mathrm{x} = \mathrm{a} \end{gathered}$$