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

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Question

(D) આપણે જાણીએ છીએ કે $	an ^{-1}(1) = \frac{\pi}{4}$.આપણે એ પણ જાણીએ છીએ કે $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,તેથી $\cos ^{-1}\left(-\frac{1}{2}

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(D) આપણે જાણીએ છીએ કે $	an ^{-1}(1) = \frac{\pi}{4}$.આપણે એ પણ જાણીએ છીએ કે $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,તેથી $\cos ^{-1}\left(-\frac{1}{2}ight) = \pi - \cos ^{-1}\left(\frac{1}{2}ight) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.આપણે જાણીએ છીએ કે $\sin ^{-1}(-x) = -\sin ^{-1}(x)$,તેથી $\sin ^{-1}\left(-\frac{1}{2}ight) = -\sin ^{-1}\left(\frac{1}{2}ight) = -\frac{\pi}{6}$.આ કિંમતોનો સરવાળો કરતા:$\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.

$\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ નું મૂલ્ય શોધો.

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