Q62 / 150 · MHT CET 2024 (09 May Shift 2) · Mathematics

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

  • A
    $\frac{\pi}{2}$
  • B
    $\frac{5 \pi}{4}$
  • C
    $\frac{7 \pi}{4}$
  • Correct answer
    D
    $\frac{3 \pi}{2}$

Solution

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

Hence, the correct answer is option: 'D'.

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