Question

Solve the following equation for $x$.
${{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x$

Answer 1

Hint: For solving this question we will use the formula ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}x-{{\tan }^{-1}}y$ to write ${{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)={{\tan }^{-1}}1-{{\tan }^{-1}}x$ directly. After that, we will find the value of ${{\tan }^{-1}}x$ and use the formula ${{\tan }^{-1}}1=\dfrac{\pi }{4}$ and $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ for calculating the suitable value of $x$.

Complete step-by-step solution -
Given:
We have to find a suitable value of $x$ and we have following equation:
${{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x$
Now, before we proceed we should know the following formula:
$\begin{align}
 &{{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}x-{{\tan }^{-1}}y\text{ }\left( \text{if }xy>-1 \right).............................\left( 1 \right) \\
 & {{\tan }^{-1}}1=\dfrac{\pi }{4}..................................\left( 2 \right) \\
 & \tan \left( {{\tan }^{-1}}x \right)= x\ \text{where}\ x \in \text{R} ..........................\left( 3 \right) \\
 & \tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}.................................\left( 4 \right) \\
\end{align} $
Now, we will use the formula from the equation (1) to write ${{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)={{\tan }^{-1}}1-{{\tan }^{-1}}x$ . Then,
$\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x \\
 & \Rightarrow {{\tan }^{-1}}1-{{\tan }^{-1}}x=\dfrac{1}{2}{{\tan }^{-1}}x \\
 & \Rightarrow {{\tan }^{-1}}1={{\tan }^{-1}}x+\dfrac{1}{2}{{\tan }^{-1}}x \\
 & \Rightarrow {{\tan }^{-1}}1=\dfrac{3}{2}{{\tan }^{-1}}x \\
\end{align} $
Now, we will use the formula from the equation (2) to write ${{\tan }^{-1}}1=\dfrac{\pi }{4}$ in the above equation. Then,
$\begin{align}
  & {{\tan }^{-1}}1=\dfrac{3}{2}{{\tan }^{-1}}x \\
 & \Rightarrow \dfrac{\pi }{4}=\dfrac{3}{2}{{\tan }^{-1}}x \\
 & \Rightarrow \dfrac{3}{2}{{\tan }^{-1}}x=\dfrac{\pi }{4} \\
 & \Rightarrow {{\tan }^{-1}}x=\dfrac{\pi }{4}\times \dfrac{2}{3} \\
 & \Rightarrow {{\tan }^{-1}}x=\dfrac{\pi }{6} \\
\end{align}$
Now, we will apply $\tan $ function on both sides in the above equation. Then,
$\begin{align}
  & {{\tan }^{-1}}x=\dfrac{\pi }{6} \\
 & \Rightarrow \tan \left( {{\tan }^{-1}}x \right)=\tan \dfrac{\pi }{6} \\
\end{align}$
Now, we will use the formula from the equation (3) to write $\tan \left( {{\tan }^{-1}}x \right)=x$ where $x \in \text{R} $ in the above equation and formula from the equation (4) to write $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ in the above equation. Then,
$\begin{align}
  & \tan \left( {{\tan }^{-1}}x \right)=\tan \dfrac{\pi }{6} \\
 & \Rightarrow x=\dfrac{1}{\sqrt{3}} \\
\end{align}$
Now, from the above result, it is evident that, $\dfrac{1}{\sqrt{3}}>-1$ so, we can write ${{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)={{\tan }^{-1}}1-{{\tan }^{-1}}x$ . Then, the value of $x$ will be equal to $\dfrac{1}{\sqrt{3}}$ .
Thus, if ${{\tan }^{-1}}\left( \dfrac{1-x}{1+x} \right)=\dfrac{1}{2}{{\tan }^{-1}}x$ then, the suitable value of $x$ will be $\dfrac{1}{\sqrt{3}}$.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct result quickly. Then, we should apply formulas of inverse trigonometric functions like ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}x-{{\tan }^{-1}}y$ in a correct manner and we should check our final answer with the condition of the formula for the justification of our answer. Moreover, we should avoid calculation mistakes while solving to get the correct result.