cos ^{-1} frac{3}{5} + sin ^{-1} frac{5}{13} | vedclass

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(A) We know that $\cos ^{-1} \left(\frac{3}{5}ight) = 	an ^{-1} \left(\frac{4}{3}ight)$ and $\sin ^{-1} \left(\frac{5}{13}ight) = 	an ^{-1} \l

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) We know that $\cos ^{-1} \left(\frac{3}{5}ight) = 	an ^{-1} \left(\frac{4}{3}ight)$ and $\sin ^{-1} \left(\frac{5}{13}ight) = 	an ^{-1} \left(\frac{5}{12}ight)$.Substituting these into the expression:$	an ^{-1} \left(\frac{4}{3}ight) + 	an ^{-1} \left(\frac{5}{12}ight) + 	an ^{-1} \left(\frac{16}{63}ight)$Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left(\frac{x+y}{1-xy}ight)$ for the first two terms:$= 	an ^{-1} \left[\frac{\frac{4}{3} + \frac{5}{12}}{1 - \left(\frac{4}{3} 	imes \frac{5}{12}ight)}ight] + 	an ^{-1} \left(\frac{16}{63}ight)$$= 	an ^{-1} \left[\frac{\frac{16+5}{12}}{\frac{36-20}{36}}ight] + 	an ^{-1} \left(\frac{16}{63}ight) = 	an ^{-1} \left(\frac{21}{12} 	imes \frac{36}{16}ight) + 	an ^{-1} \left(\frac{16}{63}ight)$$= 	an ^{-1} \left(\frac{63}{16}ight) + 	an ^{-1} \left(\frac{16}{63}ight)$Since $	an ^{-1} x + 	an ^{-1} \left(\frac{1}{x}ight) = \frac{\pi}{2}$ for $x > 0$,we get:$= \frac{\pi}{2}$.

$\cos ^{-1} \frac{3}{5} + \sin ^{-1} \frac{5}{13} + \tan ^{-1} \frac{16}{63} = $

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