If $\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1} P$, then the value of $P$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & \sin ^{-1}\left(\frac{3}{5}ight)+\cos ^{-1}\left(\frac{12}{13}ight) \ & =\sin ^{-1}\left(\frac{3}{5}ight)+\sin ^{-1} \sqrt{1-\left(\frac{12}{13}ight)^2} \ & =\sin ^{-1}\left(\frac{3}{5}ight)+\sin ^{-1}\left(\frac{5}{13}ight) \ & =\sin ^{-1}\left[\frac{3}{5} \sqrt{1-\left(\frac{5}{13}ight)^2}+\frac{5}{13} \sqrt{1-\left(\frac{3}{5}ight)^2}ight] \ & =\sin ^{-1}\left(\frac{3}{5} 	imes \frac{12}{13}+\frac{5}{13} 	imes \frac{4}{5}ight) \ & =\sin ^{-1}\left(\frac{56}{65}ight)\end{aligned}$ $	herefore \quad P=\frac{56}{65}$

If $\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1} P$, then the value of…

Solution