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

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(C) Let $x = \sin^{-1}(\frac{4}{5})$,$y = \sin^{-1}(\frac{5}{13})$,and $z = \sin^{-1}(\frac{16}{65})$.Converting to $	an^{-1}$ form:$x = 	an^{-1}(\f

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

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(C) Let $x = \sin^{-1}(\frac{4}{5})$,$y = \sin^{-1}(\frac{5}{13})$,and $z = \sin^{-1}(\frac{16}{65})$.Converting to $	an^{-1}$ form:$x = 	an^{-1}(\frac{4}{3})$,$y = 	an^{-1}(\frac{5}{12})$,$z = 	an^{-1}(\frac{16}{63})$.Now,calculate $x + y = 	an^{-1}(\frac{4}{3}) + 	an^{-1}(\frac{5}{12}) = 	an^{-1}(\frac{4/3 + 5/12}{1 - (4/3)(5/12)}) = 	an^{-1}(\frac{16/12 + 5/12}{1 - 20/36}) = 	an^{-1}(\frac{21/12}{16/36}) = 	an^{-1}(\frac{7/4}{4/9}) = 	an^{-1}(\frac{63}{16})$.Then,$(x + y) + z = 	an^{-1}(\frac{63}{16}) + 	an^{-1}(\frac{16}{63})$.Since $	an^{-1}(a) + 	an^{-1}(1/a) = \frac{\pi}{2}$ for $a > 0$,we have $(x + y) + z = \frac{\pi}{2}$.Finally,$2 \pi - (x + y + z) = 2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$.

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

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