Question

If \[{{\tanh }^{-1}}\left( x+iy \right)=\dfrac{1}{2}{{\tanh }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}+{{y}^{2}}} \right)+\dfrac{i}{2}{{\tan }^{-1}}\left( \dfrac{2y}{1-{{x}^{2}}-{{y}^{2}}} \right);x,y\in R\] then \[{{\tanh }^{-1}}\left( iy \right)\] is
1)\[{{\tanh }^{-1}}y\]
2) \[-i{{\tanh }^{-1}}y\]
3) \[i{{\tan }^{-1}}y\]
4) \[-{{\tan }^{-1}}y\]

Answer 1

Hint: In this problem, we have to find the value for the given trigonometric expression. We are given a condition that x and y belong to real numbers. We can first write the given expression we can then substitute 0 for x and simplify the terms step by step. We can cancel the inverse and the trigonometric functions and simplify it to get the required answer.

Complete step by step solution:
 Here we have to find the value of the given expression.
We know that the expression given is
\[{{\tanh }^{-1}}\left( x+iy \right)=\dfrac{1}{2}{{\tanh }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}+{{y}^{2}}} \right)+\dfrac{i}{2}{{\tan }^{-1}}\left( \dfrac{2y}{1-{{x}^{2}}-{{y}^{2}}} \right);x,y\in R\]
Let \[y=\tan \theta \]
We can now substitute x = 0 in the above given expression, we get
\[\Rightarrow {{\tanh }^{-1}}\left( 0+iy \right)=\dfrac{1}{2}{{\tanh }^{-1}}\left( \dfrac{0}{1+{{y}^{2}}} \right)+\dfrac{i}{2}{{\tan }^{-1}}\left( \dfrac{2y}{1-{{y}^{2}}} \right)\]
We can now simplify the above step, we get
\[\Rightarrow {{\tanh }^{-1}}\left( iy \right)=0+\dfrac{i}{2}{{\tan }^{-1}}\left( \dfrac{2y}{1-{{y}^{2}}} \right)\]
We can now substitute (1) in the RHS, we get
\[\Rightarrow {{\tanh }^{-1}}\left( iy \right)=0+\dfrac{i}{2}{{\tan }^{-1}}\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)\]
We can now substitute \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\] in the above step, we get
\[\Rightarrow {{\tanh }^{-1}}\left( iy \right)=\dfrac{i}{2}{{\tan }^{-1}}\tan 2\theta \]
We can now cancel the tangent and the arc tangent, we get
\[\Rightarrow {{\tanh }^{-1}}\left( iy \right)=i\theta \]
From (1), we can write theta as,
\[\Rightarrow {{\tanh }^{-1}}\left( iy \right)=i{{\tan }^{-1}}y\]
Hence, the value of \[{{\tanh }^{-1}}\left( iy \right)=i{{\tan }^{-1}}y\].
Therefore, the answer is option 3) \[i{{\tan }^{-1}}y\].

Note: We should always remember that the tangent and the arctangent get cancelled as they are inverse to each other. We should always remember some of the trigonometric identities such as \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\]. We should concentrate while simplifying the terms and write the final answer. We should know that when \[y=\tan \theta \], then the theta value can be written as \[\theta ={{\tan }^{-1}}y\].