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Question: If \[{{\tanh }^{-1}}\left( x+iy \right)=\dfrac{1}{2}{{\tanh }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}+{{y...

If tanh⁡−1(x+iy)=12tanh⁡−1(2x1+x2+y2)+i2tan⁡−1(2y1−x2−y2);x,y∈R{{\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(iy){{\tanh }^{-1}}\left( iy \right) is
1)tanh⁡−1y{{\tanh }^{-1}}y
2) −itanh⁡−1y-i{{\tanh }^{-1}}y
3) itan⁡−1yi{{\tan }^{-1}}y
4) −tan⁡−1y-{{\tan }^{-1}}y

Explanation

Solution

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(x+iy)=12tanh⁡−1(2x1+x2+y2)+i2tan⁡−1(2y1−x2−y2);x,y∈R{{\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⁡θy=\tan \theta
We can now substitute x = 0 in the above given expression, we get
⇒tanh⁡−1(0+iy)=12tanh⁡−1(01+y2)+i2tan⁡−1(2y1−y2)\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
⇒tanh⁡−1(iy)=0+i2tan⁡−1(2y1−y2)\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
⇒tanh⁡−1(iy)=0+i2tan⁡−1(2tan⁡θ1−tan⁡2θ)\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⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } in the above step, we get
⇒tanh⁡−1(iy)=i2tan⁡−1tan⁡2θ\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
⇒tanh⁡−1(iy)=iθ\Rightarrow {{\tanh }^{-1}}\left( iy \right)=i\theta
From (1), we can write theta as,
⇒tanh⁡−1(iy)=itan⁡−1y\Rightarrow {{\tanh }^{-1}}\left( iy \right)=i{{\tan }^{-1}}y
Hence, the value of tanh⁡−1(iy)=itan⁡−1y{{\tanh }^{-1}}\left( iy \right)=i{{\tan }^{-1}}y.
Therefore, the answer is option 3) itan⁡−1yi{{\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⁡2θ=2tan⁡θ1−tan⁡2θ\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⁡θy=\tan \theta , then the theta value can be written as θ=tan⁡−1y\theta ={{\tan }^{-1}}y.