Q85 / 160 · TS EAMCET 2017 · Mathematics
$\sin ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} \sqrt{\frac{2}{3}}=$
- A$\sin ^{-1} \frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}$
- Correct answerB$\pi-\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$
- C$-\pi-\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$
- D$\pi+\sin ^{-1}\left(\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right)$
Solution
We have,
$\begin{aligned}
& \sin ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} \sqrt{\frac{2}{3}} \\
& \quad=\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{1-\left(\sqrt{\frac{2}{3}}\right)^2}+\sqrt{\frac{2}{3}} \sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2}\right] \\
& \text {As } 0 < \frac{\sqrt{3}}{2}=0866, \sqrt{\frac{2}{3}}=0.816 \leq 1 \text { and }
\end{aligned}$
$\begin{aligned}
\left(\frac{\sqrt{3}}{2}\right)^2+ & \left(\sqrt{\frac{2}{3}}\right)^2=1.4167>1 \\
& x \sqrt{1-y^2}+y \sqrt{1-x^2}, \text { if } 0 < x, y \leq 1 \text { and } \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{1-\frac{2}{3}}+\sqrt{\frac{2}{3}} \sqrt{1-\frac{3}{4}}\right] \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \sqrt{\frac{3-2}{3}}+\sqrt{\frac{2}{3}} \sqrt{\frac{4-3}{4}}\right] \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{1}{2}\right] \\
& =\pi-\sin ^{-1}\left[\frac{\sqrt{3}+\sqrt{2}}{2 \sqrt{3}}\right]
\end{aligned}$
Hence, the correct answer is option: 'B'.