Question

Solve for x: \[{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x,x>0\]
A. $\sqrt{3}$
B. 1
C. $-1$
D. $\dfrac{1}{\sqrt{3}}$

Answer 1

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

Complete step-by-step solution:
Let’s assume \[{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x=\alpha \]. We use the concept of inverse theorem of trigonometric functions to find \[{{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\alpha \] and \[\dfrac{1}{2}{{\tan }^{-1}}x=\alpha \].
So, \[\left( \dfrac{1-x}{1+x} \right)=\tan \alpha \] and \[x=\tan \left( 2\alpha \right)\].
We now use the concept of multiple angles to find the formula \[\tan \left( 2\alpha \right)=\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\].
We put the values and get
\[\tan \left( 2\alpha \right)=\dfrac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\Rightarrow x=\dfrac{2\left( \dfrac{1-x}{1+x} \right)}{1-{{\left( \dfrac{1-x}{1+x} \right)}^{2}}}\].
We now multiply with \[{{\left( 1+x \right)}^{2}}\] to get \[x=\dfrac{2\left( 1-x \right)\left( 1+x \right)}{{{\left( 1+x \right)}^{2}}-{{\left( 1-x \right)}^{2}}}=\dfrac{2\left( 1-{{x}^{2}} \right)}{4x}\].
Simplifying we get \[2{{x}^{2}}=\left( 1-{{x}^{2}} \right)\] which gives \[3{{x}^{2}}=1\].
We now solve the quadratic equation to get
\[\begin{align}
  & 3{{x}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}=\dfrac{1}{3} \\
 & \Rightarrow x=\pm \dfrac{1}{\sqrt{3}} \\
\end{align}\]
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 $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.