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

Question

If $\tan \mathrm{h}^{-1}(x+i y)=\frac{1}{2} \tan \mathrm{h}^{-1}\left(\frac{2 x}{1+x^2+y^2}\right)+\frac{i}{2} \tan ^{-1}\left(\frac{2 y}{1-x^2-y^2}\right)$, where $x, y \in \mathbf{R}$, then $\tan \mathrm{h}^{-1}(i y)=$, $i \tanh ^{-1}(y)$, $-i \tanh ^{-1}(y)$, $i \tan ^{-1}(y)$, $-i \tan ^{-1}(y)$

MARKS App · Accepted Answer

We have, $\tan \mathrm{h}^{-1}(x)=\frac{1}{i} \tan ^{-1}(i x)$ Put, $\quad x=i y$, we get $\tan \mathrm{h}^{-1}(i y)=\frac{1}{i} \tan ^{-1}(-y)=\frac{-1}{i} \tan ^{-1}(y)=i \tan ^{-1}(y)$ Hence, option (c) is correct.

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