Question

If $\tan h^{-1}\left(x+iy\right)=\frac{1}{2}\tan h^{-1}\left(\frac{2x}{1+x^2+y^2}\right)+\frac{i}{2}{\tan }^{-1}\left(\frac{2y}{1-x^2-y^2}\right);x,y\in R$ then $\tan h^{-1}\left(iy\right)$ is

• A. $i\tan h^{-1}\left(y\right)$
• B. $-i\tan h^{-1}\left(y\right)$
• C. $i{\tan }^{-1}\left(y\right)$
• D. $-{\tan }^{-1}\left(y\right)$

Answer 1

The correct option is C

$i{\tan }^{-1}\left(y\right)$

Explanation for the correct option

Given that $\tan h^{-1}\left(x+iy\right)=\frac{1}{2}\tan h^{-1}\left(\frac{2x}{1+x^2+y^2}\right)+\frac{i}{2}{\tan }^{-1}\left(\frac{2y}{1-x^2-y^2}\right)$

Substituting $x=0$, we get,

$\tan h^{-1}\left(iy\right)=\frac{i}{2}{\tan }^{-1}\left(\frac{2y}{1-y^2}\right)$

Say $y=\tan \left(\theta \right)$

$\begin{array}{crcll}\Rightarrow & \tan h^{-1}\left(iy\right)& =& \frac{i}{2}{\tan }^{-1}\left(\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}\right)& \\ & & =& \frac{i}{2}{\tan }^{-1}\left(\tan \left(2\theta \right)\right)& \left[\because \frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}=\tan \left(2\theta \right)\right]\\ & & =& \frac{i}{2}\left(2\theta \right)& \\ & & =& i\theta & \\ \Rightarrow & \tan h^{-1}\left(iy\right)& =& i{\tan }^{-1}\left(y\right)& \left[\because y=\tan \left(\theta \right)\right]\end{array}$

Hence, the correct option is option(C) i.e. $i{\tan }^{-1}\left(y\right)$