Question

$\tan h^{-1}x=$

• A. $\frac{1}{2}\log \left(\frac{x+1}{x-1}\right)$
• B. $\frac{1}{2}\log \left(\frac{x-1}{x+1}\right)$
• C. $\frac{1}{2}\log \left(\frac{1-x}{1+x}\right)$
• D. $\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$

Answer 1

The correct option is D

$\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$

Explanation for the correct answer

Solving the given trigonometric expression

Given, ${\tanh }^{-1}\mathrm{x}$

Therefore,

$$\begin{gathered} y=\tan h^{-1}x \\ \Rightarrow x=\tan hy \\ =\frac{e^y-e^{-y}}{e^y+e^{-y}}\left[\because \tan hy=\frac{e^y-e^{-y}}{e^y+e^{-y}}\right] \\ =\frac{e^{2y}-1}{e^{2y}+1} \end{gathered}$$

This gives:

$$\begin{gathered} x\left(e^{2y}+1\right)=e^{2y}-1 \\ \Rightarrow \left(x-1\right)e^{2y}+x+1=0 \\ \Rightarrow e^{2y}=\frac{-(1+x)}{x-1} \\ =\frac{1+x}{1-x} \end{gathered}$$

Taking logarithm on both sides:

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

Therefore, the value of ${\tanh }^{-1}\mathrm{x}$ is $\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$.

Hence, the correct answer is Option (D).