If sin ^{-1} frac{alpha}{17}+cos ^{-1} frac{4 | vedclass

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(A) Given $\cos ^{-1} \frac{4}{5} = 	an ^{-1} \frac{3}{4}$.Substituting this into the equation: $\sin ^{-1} \frac{\alpha}{17} = 	an ^{-1} \frac{77}{

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$\pi$

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(A) Given $\cos ^{-1} \frac{4}{5} = 	an ^{-1} \frac{3}{4}$.Substituting this into the equation: $\sin ^{-1} \frac{\alpha}{17} = 	an ^{-1} \frac{77}{36} - 	an ^{-1} \frac{3}{4}$.Using the formula $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \frac{x-y}{1+xy}$:$\sin ^{-1} \frac{\alpha}{17} = 	an ^{-1} \left( \frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}} ight) = 	an ^{-1} \left( \frac{\frac{77-27}{36}}{1 + \frac{231}{144}} ight) = 	an ^{-1} \left( \frac{50/36}{375/144} ight) = 	an ^{-1} \left( \frac{50}{36} \cdot \frac{144}{375} ight) = 	an ^{-1} \left( \frac{200}{375} ight) = 	an ^{-1} \frac{8}{15}$.Since $	an ^{-1} \frac{8}{15} = \sin ^{-1} \frac{8}{17}$,we have $\sin ^{-1} \frac{\alpha}{17} = \sin ^{-1} \frac{8}{17}$,which implies $\alpha = 8$.Now,we evaluate $\sin ^{-1}(\sin 8) + \cos ^{-1}(\cos 8)$.Since $2\pi $\sin ^{-1}(\sin 8) = 8 - 2\pi$.Since $2\pi $\cos ^{-1}(\cos 8) = 8 - 2\pi$ is incorrect; rather,since $2\pi Thus,$\cos ^{-1}(\cos 8) = 8 - 2\pi$.Therefore,$\sin ^{-1}(\sin 8) + \cos ^{-1}(\cos 8) = (8 - 2\pi) + (8 - 2\pi) = 16 - 4\pi$ is not correct. Let's re-evaluate: $\sin ^{-1}(\sin 8) = 8 - 2\pi$ and $\cos ^{-1}(\cos 8) = 8 - 2\pi$ is wrong. Actually,$2\pi \approx 6.28$. $8$ is in the interval $(2\pi, 3\pi)$.$\sin ^{-1}(\sin 8) = 8 - 2\pi$.$\cos ^{-1}(\cos 8) = 8 - 2\pi$ is wrong. $\cos(8) = \cos(8 - 2\pi)$. Since $8 - 2\pi \approx 1.72$,which is in $[0, \pi]$,$\cos ^{-1}(\cos 8) = 8 - 2\pi$.Sum $= (8 - 2\pi) + (8 - 2\pi) = 16 - 4\pi$. Wait,let's re-check the options. If the answer is $\pi$,then $\sin ^{-1}(\sin 8) = 3\pi - 8$ and $\cos ^{-1}(\cos 8) = 8 - 2\pi$. Sum $= \pi$.

If $\sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0$ and $0 < \alpha < 13$,then $\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha)$ is equal to $.........$.

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