Question

The value of
$\sin ^{-1} (\frac {2 \sqrt {2}}{3})+\sin^{-1} (\frac {1}{3}) $ is equal to

• A. $\frac {\pi} {6}$
• B. $\frac {\pi} {2}$
• C. $\frac {\pi} {4}$
• D. $\frac {2\pi} {4}$

Answer 1

Answer: (B)

$\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3} $
$=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\cos ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^{2}} $
$=\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right) $
$=\frac{\pi}{2} \left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$