If {tan ^{ - 1}}x + 2{cot ^{ - 1}}x = frac{{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: ${	an ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ We know that ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$ for all

Vedclass · Accepted Answer

$\sqrt 3 $

Vedclass · Answer

(C) Given equation: ${	an ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ We know that ${	an ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$ for all $x \in \mathbb{R}$. We can rewrite the given equation as: $({	an ^{ - 1}}x + {\cot ^{ - 1}}x) + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ Substituting the identity: $\frac{\pi }{2} + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$ Subtracting $\frac{\pi }{2}$ from both sides: ${\cot ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{2}$ ${\cot ^{ - 1}}x = \frac{{4\pi - 3\pi }}{6} = \frac{\pi }{6}$ Taking the cotangent of both sides: $x = \cot \left( \frac{\pi }{6} ight)$ Since $\cot \left( \frac{\pi }{6} ight) = \sqrt 3 $,we get $x = \sqrt 3 $.

If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$

Similar Questions

The value of $\operatorname{cosec}^{-1}(\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)-\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is equal to

If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{\pi}{2}$,then the value of $x^2 + y^2 + z^2 + 2xyz$ is equal to

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$,then $x$ =

$\cos \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\right]=$

If $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$,then the value of $\cos ^{-1} x+\sin ^{-1} y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module