કિંમત શોધો: sin^{-1}(12/13) + cos^{-1}(4/5) + | vedclass

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Question

(C) ધારો કે $\alpha = \sin^{-1}(12/13)$, તો $\sin \alpha = 12/13$। તેથી, $	an \alpha = 12/5$।ધારો કે $\beta = \cos^{-1}(4/5)$, તો $\cos \beta = 4/5$।

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(C) ધારો કે $\alpha = \sin^{-1}(12/13)$, તો $\sin \alpha = 12/13$। તેથી, $	an \alpha = 12/5$।ધારો કે $\beta = \cos^{-1}(4/5)$, તો $\cos \beta = 4/5$। તેથી, $	an \beta = 3/4$।હવે, $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{12/5 + 3/4}{1 - (12/5)(3/4)} = \frac{(48+15)/20}{1 - 36/20} = \frac{63/20}{-16/20} = -63/16$।કારણ કે $\alpha + \beta$ બીજા ચરણમાં છે, $\alpha + \beta = \pi - 	an^{-1}(63/16)$।તેથી, $\alpha + \beta + 	an^{-1}(63/16) = \pi - 	an^{-1}(63/16) + 	an^{-1}(63/16) = \pi$।

કિંમત શોધો: $\sin^{-1}(12/13) + \cos^{-1}(4/5) + \tan^{-1}(63/16) = $

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