Find the principal value of cot ^{-1}left(fra | vedclass

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

Question

(B) Let $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right) = y$. Then,$\cot y = \frac{-1}{\sqrt{3}}$.Since $\cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}ight) = y$. Then,$\cot y = \frac{-1}{\sqrt{3}}$.Since $\cot \left(\frac{\pi}{3}ight) = \frac{1}{\sqrt{3}}$,we have $\cot y = -\cot \left(\frac{\pi}{3}ight)$.Using the property $\cot(\pi - 	heta) = -\cot 	heta$,we get $\cot y = \cot \left(\pi - \frac{\pi}{3}ight) = \cot \left(\frac{2\pi}{3}ight)$.The range of the principal value branch of $\cot^{-1}$ is $(0, \pi)$.Since $\frac{2\pi}{3} \in (0, \pi)$,the principal value is $\frac{2\pi}{3}$.

Find the principal value of $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.

Similar Questions

Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $a=3, b=4$ and $A=\sin^{-1}\left(\frac{3}{4}\right)$,then the angle $B$ is (in $^{\circ}$)

Considering only the principal values of an inverse function,the set $A = \{x \geq 0 \mid \tan^{-1} x + \tan^{-1} 6x = \frac{\pi}{4}\}$

If ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta $,then $\tan \theta =$

The value of $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right)$ is ....... $^o$.

Solve $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module