If \(\tan \mathrm{h}^{-1}(x+i y)=\frac{1}{2} \tan \mathrm{h}^{-1}\left(\frac{2…
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 \(\tan \mathrm{h}^{-1}(i y)=\)
\(i \tanh ^{-1}(y)\)
\(-i \tanh ^{-1}(y)\)
\(i \tan ^{-1}(y)\)
\(-i \tan ^{-1}(y)\)
Solution
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.