sin left[ frac{pi }{2} - sin^{-1} left( -frac | vedclass

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

Question

(C) We know that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,which implies $\frac{\pi}{2} - \sin^{-1}(x) = \cos^{-1}(x)$.Substituting $x = -\frac{\s

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) We know that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,which implies $\frac{\pi}{2} - \sin^{-1}(x) = \cos^{-1}(x)$.Substituting $x = -\frac{\sqrt{3}}{2}$,we get:$\sin \left[ \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \right]$Since $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,we have:$\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \pi - \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Now,$\sin \left( \frac{5\pi}{6} \right) = \sin \left( \pi - \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$.

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

Similar Questions

The range of the function $f(x) = \sqrt{|\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||}$ is

Solve $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1} \left(\frac{x-y}{x+y}\right)$ is equal to

Let the maximum value of $(\sin^{-1}x)^{2} + (\cos^{-1}x)^{2}$ for $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n}\pi^{2}$, where $\gcd(m, n) = 1$. Then $m+n$ is equal to ........... .

If $k \le \sin^{-1}x + \cos^{-1}x + \tan^{-1}x \le K$,then

The value of $\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module