Question

Solve the following equation:
${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$

Answer 1

- Hint: In the given equation, convert the left hand side of the equation in terms of ${{\cot }^{-1}}x$ using the trigonometric inverse identity ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$. Then solve the equation and find the value of ${{\cot }^{-1}}x$ and then take cot on both sides to get the value of x.

Complete step-by-step solution -

The equation given in the question is:
${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$
Rewriting the above equation as:
${{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\cot }^{-1}}x=\dfrac{2\pi }{3}$
We know from the inverse trigonometric identity that ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ so substituting this value of ${{\tan }^{-1}}x+{{\cot }^{-1}}x$ in the above equation we get,
\[\begin{align}
  & \dfrac{\pi }{2}+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\
 & \Rightarrow {{\cot }^{-1}}x=\dfrac{2\pi }{3}-\dfrac{\pi }{2} \\
 & \Rightarrow {{\cot }^{-1}}x=\dfrac{4\pi -3\pi }{6} \\
 & \Rightarrow {{\cot }^{-1}}x=\dfrac{\pi }{6} \\
\end{align}\]
Now, taking cot on both the sides of the equation we get,
$\cot \left( {{\cot }^{-1}}x \right)=\cot \dfrac{\pi }{6}$
We know from the trigonometric ratios that $\cot \dfrac{\pi }{6}=\sqrt{3}$ so substituting this value in the above equation we get,
$x=\sqrt{3}$
Hence, the value of x in the given equation is $x=\sqrt{3}$.

Note: You can verify the value of x that you are getting above by substituting the value of x in the given equation ${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$ and then check whether after substituting the value of x in the given equation, you are getting left hand side equal to right hand side or not.
${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$
Solving L.H.S of the above equation we get,
${{\tan }^{-1}}\left( \sqrt{3} \right)+2{{\cot }^{-1}}\left( \sqrt{3} \right)$
We know from the inverse trigonometric values that:
${{\tan }^{-1}}\left( \sqrt{3} \right)=\dfrac{\pi }{3}$
${{\cot }^{-1}}\left( \sqrt{3} \right)=\dfrac{\pi }{6}$
Plugging the above values in the expression ${{\tan }^{-1}}\left( \sqrt{3} \right)+2{{\cot }^{-1}}\left( \sqrt{3} \right)$ we get,
$\begin{align}
  & \dfrac{\pi }{3}+2\dfrac{\pi }{6} \\
 & =\dfrac{\pi }{3}+\dfrac{\pi }{3}=\dfrac{2\pi }{3} \\
\end{align}$
The R.H.S of the given equation is equal to $\dfrac{2\pi }{3}$.
From the above simplification of the left hand side of the given equation we get $\dfrac{2\pi }{3}$ which is equal to the right hand side R.H.S of the given equation.
So, the value of x that we have calculated above is satisfying the given equation.
Hence, the value of x that we have got above is correct.