Question

The value of $2sin^{- 1} \frac{4}{5}+2sin^{- 1} ⁡ \frac{5}{13}+2sin^{- 1} ⁡ \frac{16}{65}$ is equal to

• A. $\frac{3 \pi }{2}$
• B. $\frac{\pi }{2}$
• C. $\pi $
• D. $2\pi $

Answer 1

Answer: (C)

$2\left[\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right]$
$=2 \sin ^{-1}\left(\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}+\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right)+2 \sin ^{-1} \frac{16}{65}$
$=2 \sin ^{-1}\left(\frac{48}{65}+\frac{15}{65}\right)+2 \sin ^{-1}\left(\frac{16}{65}\right)$
$=2\left[\sin ^{-1}\left(\frac{63}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right]$
$=2\left[\cos ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right]\left(\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}\right)$
$=\pi$