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

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Question

(C) हम जानते हैं कि $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,जिसका अर्थ है कि $\frac{\pi}{2} - \sin^{-1}(x) = \cos^{-1}(x)$.$x = -\frac{\sqrt{3}}

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) हम जानते हैं कि $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,जिसका अर्थ है कि $\frac{\pi}{2} - \sin^{-1}(x) = \cos^{-1}(x)$.$x = -\frac{\sqrt{3}}{2}$ रखने पर,हमें प्राप्त होता है:$\sin \left[ \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \right]$चूंकि $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,इसलिए:$\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \pi - \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.अब,$\sin \left( \frac{5\pi}{6} \right) = \sin \left( \pi - \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$.

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

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