sin ^{-1}left(frac{1}{2}right)+cos ^{-1}left( | vedclass

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(B) આપણે જાણીએ છીએ કે $\sin ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (અથવા $30^{\circ}$).આપણે જાણીએ છીએ કે $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\r

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$\pi$

Vedclass · Answer

(B) આપણે જાણીએ છીએ કે $\sin ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (અથવા $30^{\circ}$).આપણે જાણીએ છીએ કે $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ (અથવા $30^{\circ}$).આપણે જાણીએ છીએ કે $\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \pi - \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (અથવા $120^{\circ}$).આ કિંમતોનો સરવાળો કરતા:$\frac{\pi}{6} + \frac{\pi}{6} + \frac{2\pi}{3} = \frac{2\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{3} + \frac{2\pi}{3} = \frac{3\pi}{3} = \pi$.

$\sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=$

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