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Question: Solve for x: \[{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x,x>0\] A....

Solve for x: tan⁡−1(1−x1+x)=12tan⁡−1x,x>0{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x,x>0
A. 3\sqrt{3}
B. 1
C. −1-1
D. 13\dfrac{1}{\sqrt{3}}

Explanation

Solution

We first try to assume the variable of the given equation where tan⁡−1(1−x1+x)=12tan⁡−1x=α{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x=\alpha . Using the theorem of inverse, we get the values for the (1−x1+x)=tan⁡α\left( \dfrac{1-x}{1+x} \right)=\tan \alpha and x=tan⁡(2α)x=\tan \left( 2\alpha \right). Then we use the multiple angles to find the formula tan⁡(2α)=2tan⁡α1−tan⁡2α\tan \left( 2\alpha \right)=\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }. We put the values and find the solution for x.

Complete step-by-step solution:
Let’s assume tan⁡−1(1−x1+x)=12tan⁡−1x=α{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x=\alpha . We use the concept of inverse theorem of trigonometric functions to find tan⁡−1(1−x1+x)=α{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\alpha and 12tan⁡−1x=α\dfrac{1}{2}{{\tan }^{-1}}x=\alpha .
So, (1−x1+x)=tan⁡α\left( \dfrac{1-x}{1+x} \right)=\tan \alpha and x=tan⁡(2α)x=\tan \left( 2\alpha \right).
We now use the concept of multiple angles to find the formula tan⁡(2α)=2tan⁡α1−tan⁡2α\tan \left( 2\alpha \right)=\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }.
We put the values and get
tan⁡(2α)=2tan⁡α1−tan⁡2α⇒x=2(1−x1+x)1−(1−x1+x)2\tan \left( 2\alpha \right)=\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\Rightarrow x=\dfrac{2\left( \dfrac{1-x}{1+x} \right)}{1-{{\left( \dfrac{1-x}{1+x} \right)}^{2}}}.
We now multiply with (1+x)2{{\left( 1+x \right)}^{2}} to get x=2(1−x)(1+x)(1+x)2−(1−x)2=2(1−x2)4xx=\dfrac{2\left( 1-x \right)\left( 1+x \right)}{{{\left( 1+x \right)}^{2}}-{{\left( 1-x \right)}^{2}}}=\dfrac{2\left( 1-{{x}^{2}} \right)}{4x}.
Simplifying we get 2x2=(1−x2)2{{x}^{2}}=\left( 1-{{x}^{2}} \right) which gives 3x2=13{{x}^{2}}=1.
We now solve the quadratic equation to get

& 3{{x}^{2}}=1 \\\ & \Rightarrow {{x}^{2}}=\dfrac{1}{3} \\\ & \Rightarrow x=\pm \dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ **Now it’s given that the $$x>0$$ which means $$x=\dfrac{1}{\sqrt{3}}$$. The correct option is D.** **Note:** Although for elementary knowledge the principal domain is enough to solve. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi +a$ for $\tan \left( x \right)=\tan a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.