2 pi - left(sin ^{-1} frac{4}{5} + sin ^{-1} | vedclass

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(D) Let $S = \sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}$.First,convert $\sin ^{-1} \frac{4}{5}$ and $\sin ^{-1} \frac

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$\frac{3 \pi}{2}$

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(D) Let $S = \sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}$.First,convert $\sin ^{-1} \frac{4}{5}$ and $\sin ^{-1} \frac{5}{13}$ to $	an ^{-1}$ form:$\sin ^{-1} \frac{4}{5} = 	an ^{-1} \frac{4}{3}$ and $\sin ^{-1} \frac{5}{13} = 	an ^{-1} \frac{5}{12}$.Now,calculate the sum of the first two terms:$	an ^{-1} \frac{4}{3} + 	an ^{-1} \frac{5}{12} = 	an ^{-1} \left(\frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} 	imes \frac{5}{12}}ight) = 	an ^{-1} \left(\frac{\frac{16+5}{12}}{\frac{36-20}{36}}ight) = 	an ^{-1} \left(\frac{21}{12} 	imes \frac{36}{16}ight) = 	an ^{-1} \left(\frac{63}{16}ight)$.Since $	an ^{-1} \frac{63}{16} = \cot ^{-1} \frac{16}{63}$,we can convert this to $\sin ^{-1}$ form as $\sin ^{-1} \frac{63}{65}$.Alternatively,note that $	an ^{-1} \frac{63}{16} = \cos ^{-1} \frac{16}{65}$.Thus,$S = \cos ^{-1} \frac{16}{65} + \sin ^{-1} \frac{16}{65}$.Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get $S = \frac{\pi}{2}$.Finally,the expression is $2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$.

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

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