Question

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

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

Answer 1

Answer: (C)

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