cos ^{-1}left(cos frac{13 pi}{6}right)+tan ^{ | vedclass

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

Question

(B) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$ and $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.First,consider $\c

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(B) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$ and $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.First,consider $\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight)$.Since $\frac{13 \pi}{6} = 2 \pi + \frac{\pi}{6}$,we have $\cos \frac{13 \pi}{6} = \cos \frac{\pi}{6}$.Thus,$\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight) = \cos ^{-1}\left(\cos \frac{\pi}{6}ight) = \frac{\pi}{6}$.Next,consider $	an ^{-1}\left(	an \frac{7 \pi}{6}ight)$.Since $\frac{7 \pi}{6} = \pi + \frac{\pi}{6}$,we have $	an \frac{7 \pi}{6} = 	an \frac{\pi}{6}$.Thus,$	an ^{-1}\left(	an \frac{7 \pi}{6}ight) = 	an ^{-1}\left(	an \frac{\pi}{6}ight) = \frac{\pi}{6}$.Adding these results,we get $\frac{\pi}{6} + \frac{\pi}{6} = \frac{2 \pi}{6} = \frac{\pi}{3}$.

$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=$ . . . . . . .

Similar Questions

The value of $\sin^{-1}\left(\sin \frac{7 \pi}{6}\right)$ is equal to . . . . . . .

$A$ positive acute angle is divided into two parts whose tangents are $\frac{1}{2}$ and $\frac{1}{3}$. Then the angle is

The principal value of ${\sin ^{ - 1}}\left[ {\sin \left( {\frac{{2\pi }}{3}} \right)} \right]$ is

If $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $0 < x < 1$ and $0 < y < \frac{\pi}{2}$,then $\frac{dy}{dx} = $

The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} a$ has a solution

Vedclass Test Series

Exam Paper Generator

Online Exam Module