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 value of $\sin^{-1}\left(\sin \frac{5\pi}{3}\right)$ is equal to . . . . . . .

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

$2 \tanh^{-1} \frac{1}{2}$ is equal to

Evaluate the expression: $sec^2(\tan^{-1} 3) - \tan^2(sec^{-1} 3) = $

If the range of $\operatorname{sech}^{-1} x + \operatorname{cosech}^{-1} x$ is $[a, b]$, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module