Question

Solve for x the given inverse trigonometric equation ${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$.

Answer 1

Hint: For solving this question we will use one of the important formula from inverse trigonometric functions, i.e. ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ . After that, we will find the value of ${{\cot }^{-1}}x$ and then, we will apply trigonometric function $\cot $ and use the formula $\cot \left( {{\cot }^{-1}}x \right)=x$ where $x \in \text{R} $ and result $\cot \dfrac{\pi }{6}=\sqrt{3}$ to get the suitable value of $x$ easily.

Complete step-by-step solution -
Given:
We have to find a suitable value of $x$ and we have the following equation:
${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$
Now, before we proceed we should know the following formulas:
$\begin{align}
&{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\text{ }\left( \text{for all }x\in R \right).....................\left( 1 \right) \\
 & \cot \left( {{\cot }^{-1}}x \right)=x\ \text{where}\ x\in \text{R} ....................................................\left( 2 \right) \\
 & \cot \dfrac{\pi }{6}=\sqrt{3}............................................................\left( 3 \right) \\
\end{align}$
Now, we will use the above three formulas for solving this question.
We have the following equation:
$\begin{align}
  & {{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\
 & \Rightarrow {{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\
\end{align}$
Now, we will use the formula from the equation (1) to write ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ in the above equation. Then,
$\begin{align}
  & {{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\
 & \Rightarrow \dfrac{\pi }{2}+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\
 & \Rightarrow {{\cot }^{-1}}x=\dfrac{2\pi }{3}-\dfrac{\pi }{2} \\
 & \Rightarrow {{\cot }^{-1}}x=\dfrac{\pi }{6} \\
\end{align}$
Now, we will apply the trigonometric function $\cot $ on both sides in the above equation. Then,
$\begin{align}
  & {{\cot }^{-1}}x=\dfrac{\pi }{6} \\
 & \Rightarrow \cot \left( {{\cot }^{-1}}x \right)=\cot \dfrac{\pi }{6} \\
\end{align}$
Now, we will use the formula from the equation (2) to write $\cot \left( {{\cot }^{-1}}x \right)=x$ where $x \in \text{R} $ and formula from the equation (3) to write $\cot \dfrac{\pi }{6}=\sqrt{3}$ in the above equation. Then,
$\begin{align}
  & \cot \left( {{\cot }^{-1}}x \right)=\cot \dfrac{\pi }{6} \\
 & \Rightarrow x=\sqrt{3} \\
\end{align}$
Now, from the above result, we conclude that a suitable value of $x$ will be equal to $\sqrt{3}$ .
Thus, if ${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$ then, the suitable value of $x$ will be equal to $\sqrt{3}$ .

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, we should remember the formula ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ and apply it in such questions, to solve quickly without any tough calculation. And though the question is very easy, we should avoid calculation mistakes while solving the question.