MCQ
The value of $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ :
- A$\frac{\pi}{2}$
- B$\frac{\pi}{3}$
- ✓$\frac{2 \pi}{3}$
- D$\pi$
✓
Answer
Correct option: C.
$\frac{2 \pi}{3}$
(C)
The range of principal value branch of $\sin ^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and the range of principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.
$\therefore \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$
Hence, $\frac{\pi}{3}+2 \times \frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$
Hence correct option is (C)
The range of principal value branch of $\sin ^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and the range of principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.
$\therefore \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$
Hence, $\frac{\pi}{3}+2 \times \frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$
Hence correct option is (C)
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