cos left[sin ^{-1}left(frac{3}{5}right)+cos ^ | vedclass

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(C) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$ and $\beta = \cos^{-1}\left(\frac{12}{13}\right)$.Then $\sin \alpha = \frac{3}{5}$,which implies

Vedclass · Accepted Answer

$\frac{33}{65}$

Vedclass · Answer

(C) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$ and $\beta = \cos^{-1}\left(\frac{12}{13}\right)$.Then $\sin \alpha = \frac{3}{5}$,which implies $\cos \alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$.And $\cos \beta = \frac{12}{13}$,which implies $\sin \beta = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13}$.Using the identity $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:$\cos(\alpha + \beta) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) - \left(\frac{3}{5}\right) \left(\frac{5}{13}\right)$$= \frac{48}{65} - \frac{15}{65}$$= \frac{33}{65}$.

$\cos \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\right]=$

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