Question

If the value of ${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$, then the value of x is-
 A. $\sqrt{2}$
B. 3
C. $\sqrt{3}$
D. $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

Answer 1

As we know from property of inverse trigonometric function we can write
${{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x$ .Use this information to solve the question.

Complete step-by-step answer:
Given equation is ${{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$
Now from property of inverse trigonometric function we can write
${{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x$
On substituting value of ${{\cot }^{-1}}x$ in given equation
$\Rightarrow {{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}$
$\Rightarrow {{\tan }^{-1}}x+2\left( \dfrac{\pi }{2}-{{\tan }^{-1}}x \right)=\dfrac{2\pi }{3}$
$\Rightarrow {{\tan }^{-1}}x+2\times \dfrac{\pi }{2}-2{{\tan }^{-1}}x=\dfrac{2\pi }{3}$
$\Rightarrow \pi -{{\tan }^{-1}}x=\dfrac{2\pi }{3}$
.$\Rightarrow {{\tan }^{-1}}x=\pi -\dfrac{2\pi }{3}$
$\Rightarrow {{\tan }^{-1}}x=\dfrac{3\pi -2\pi }{3}$
$\Rightarrow {{\tan }^{-1}}x=\dfrac{\pi }{3}$
$\Rightarrow x=\tan \left( \dfrac{\pi }{3} \right)$ $\left\{ \begin{align}
  & If\,{{\tan }^{-1}}(x)=\theta \\
 & \Rightarrow x=\tan (\theta ) \\
\end{align} \right\}$
$\Rightarrow x=\sqrt{3}$
Hence option C is correct.

Note: In the above question we need to remember that we will change only one inverse trigonometric function in terms to another.
As we put value of ${{\cot }^{-1}}x$in above equation in terms of ${{\tan }^{-1}}x$and solve equation in terms of ${{\tan }^{-1}}x$. Similarly we can also write value of ${{\tan }^{-1}}x$ in terms of ${{\cot }^{-1}}x$ and then solve equation in terms of ${{\cot }^{-1}}x$.