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

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Question

(C) Let $\alpha = \sin^{-1}(12/13)$, then $\sin \alpha = 12/13$. Thus, $	an \alpha = 12/5$.Let $\beta = \cos^{-1}(4/5)$, then $\cos \beta = 4/5$. Thu

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$\pi$

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(C) Let $\alpha = \sin^{-1}(12/13)$, then $\sin \alpha = 12/13$. Thus, $	an \alpha = 12/5$.Let $\beta = \cos^{-1}(4/5)$, then $\cos \beta = 4/5$. Thus, $	an \beta = 3/4$.Now, $	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$.Since $\alpha + \beta$ is in the second quadrant, $\alpha + \beta = \pi - 	an^{-1}(63/16)$.Therefore, $\alpha + \beta + 	an^{-1}(63/16) = \pi - 	an^{-1}(63/16) + 	an^{-1}(63/16) = \pi$.

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

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