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

$\pi+\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right)$ is equal to
  1. $\frac{\pi}{2}$
  2. $\frac{5 \pi}{4}$
  3. $\frac{3 \pi}{2}$
  4. $\frac{7 \pi}{4}$

Solution

$\begin{aligned} & \pi+\left[\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =\pi+\left[\left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =\pi+\left[\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =\pi+\left[\tan ^{-1}\left(\frac{63}{16}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =\pi+\left[\cos ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =\pi+\frac{\pi}{2} \\ & =\frac{3 \pi}{2}\end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 2)

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