Question

If $\cos ^{-1} x=\alpha,(0 < x < 1)$ and $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=\frac{2 \pi}{3}$, then $\tan ^{-1}(2 x)$ equals

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (C)

$\cos ^{-1} x=\alpha \Rightarrow x=\cos \alpha$
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=\frac{2 \pi}{3}$
$2 \alpha+2 \alpha=\frac{2 \pi}{3}$
$\Rightarrow 4 \alpha=\frac{2 \pi}{3}$
$\Rightarrow \alpha=\frac{\pi}{6}$
$x=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$
$\tan ^{-1}(2 x)=\tan ^{-1}(\sqrt{3})=\pi / 3$