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AP EAMCET · Maths · Inverse Trigonometric Functions

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 \(\tan \mathrm{h}^{-1}(i y)=\)

  1. A \(i \tanh ^{-1}(y)\)
  2. B \(-i \tanh ^{-1}(y)\)
  3. C \(i \tan ^{-1}(y)\)
  4. D \(-i \tan ^{-1}(y)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(i \tan ^{-1}(y)\)

Step-by-step Solution

Detailed explanation

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.