sin ^{-1}left[cos left(sin ^{-1} frac{sqrt{3} | vedclass

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

Question

(B) We know that $\sin^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ because $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.Substituti

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(B) We know that $\sin^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ because $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.Substituting this into the expression,we get $\sin^{-1} \left[\cos \left(\frac{\pi}{3}\right)\right]$.Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,the expression becomes $\sin^{-1} \left(\frac{1}{2}\right)$.We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,therefore $\sin^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$.Thus,the correct option is $B$.

$\sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]=$ . . . . . . .

Similar Questions

The interval for which ${\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}$ holds is:

If $\sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6}$, then $x=$

The value of $x$ which satisfies the equation $\tan^{-1}x = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$ is

The value of $\sin \left(2 \sin ^{-1} 0.8\right)$ is equal to

Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module