A value of tan^{-1} left( sin left( cos^{-1} | vedclass

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

Question

(D) Let $	heta = \cos^{-1} \sqrt{\frac{2}{3}}$.Then $\cos 	heta = \sqrt{\frac{2}{3}}$.We know that $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.Substitu

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(D) Let $	heta = \cos^{-1} \sqrt{\frac{2}{3}}$.Then $\cos 	heta = \sqrt{\frac{2}{3}}$.We know that $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.Substituting the value of $\cos 	heta$,we get $\sin 	heta = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.Now,the expression becomes $	an^{-1} (\sin 	heta) = 	an^{-1} \left( \frac{1}{\sqrt{3}} ight)$.Since $	an \left( \frac{\pi}{6} ight) = \frac{1}{\sqrt{3}}$,we have $	an^{-1} \left( \frac{1}{\sqrt{3}} ight) = \frac{\pi}{6}$.

$A$ value of $\tan^{-1} \left( \sin \left( \cos^{-1} \sqrt{\frac{2}{3}} \right) \right)$ is

Similar Questions

The principal value of $\cos^{-1}[\cos(-680^{\circ})]$ is equal to: . . . . . . .

Find the value of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$.

$2 \coth^{-1}(4) + \text{sech}^{-1}\left(\frac{3}{5}\right) = $

$A$ value of $x$ for which $\sin(\cot^{-1}(1 + x)) = \cos(\tan^{-1}x)$ is

If $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$,then $\sinh \alpha=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module