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

Find the derivative: $\frac{d}{dx} \cos^{-1} \left( \frac{x - x^{-1}}{x + x^{-1}} \right)$

The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

Evaluate: $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)$

If $\cos^{-1}\left(\frac{2}{3x}\right) + \cos^{-1}\left(\frac{3}{4x}\right) = \frac{\pi}{2}$ for $x > \frac{3}{4}$,then $x$ is equal to

If $y = 2\sin^{-1} \sqrt{1-x} + \sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$,then $\frac{dy}{dx}$ equals-

Vedclass Test Series

Exam Paper Generator

Online Exam Module