The principal value of cot^{-1}left(frac{-1}{ | vedclass

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

Question

(B) Let $y = \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.Then $\cot y = \frac{-1}{\sqrt{3}}$.We know that the range of the principal value branch of $\

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(B) Let $y = \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.Then $\cot y = \frac{-1}{\sqrt{3}}$.We know that the range of the principal value branch of $\cot^{-1} x$ is $(0, \pi)$.Since $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,we have $\cot\left(\pi - \frac{\pi}{3}\right) = -\cot\left(\frac{\pi}{3}\right) = \frac{-1}{\sqrt{3}}$.Thus,$\cot\left(\frac{2\pi}{3}\right) = \frac{-1}{\sqrt{3}}$.Since $\frac{2\pi}{3} \in (0, \pi)$,the principal value is $\frac{2\pi}{3}$.

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is equal to . . . . . . .

Similar Questions

If $\frac{3x+1}{(x-1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+3}$,then $\sin^{-1} \frac{A}{B}$ is equal to

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

If $\tan \theta = - \frac{1}{\sqrt{3}}$,$\sin \theta = \frac{1}{2}$,and $\cos \theta = - \frac{\sqrt{3}}{2}$,then the principal value of $\theta$ is:

$\sin (2{\sin ^{ - 1}}0.8) = $

$\cos^{-1}\left(x^2 + \frac{1}{x^2} - 1\right) + \sin^{-1}\left(x^2 - \frac{1}{x^2}\right) + \tan^{-1}(x^2)$ is equal to (where $x \in R - \{0\}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module