If \(\cos ^{-1} x=\alpha(0 \lt x \lt 1)\) 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 \(\alpha\) is

Given equation is
\(\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)=\frac{2 \pi}{3}\)
Also, \(\cos ^{-1} x=\alpha\)
\(\Rightarrow x=\cos \alpha\)
and \(0 \lt \alpha \lt \frac{\pi}{2}\)
\(\ldots[\because 0 \lt x \lt 1]\)
Putting \(x=\cos \alpha\) in the given equation, we get
\(\sin ^{-1}\left(2 \cos \alpha \sqrt{1-\cos ^2 \alpha}\right)+\sec ^{-1}\left(\frac{1}{2 \cos ^2 \alpha-1}\right)\) \(=\frac{2 \pi}{3} \)
\( \Rightarrow \sin ^{-1}(2 \cos \alpha \sin \alpha)+\sec ^{-1}\left(\frac{1}{\cos 2 \alpha}\right)=\frac{2 \pi}{3}\)
\(\begin{aligned} & \Rightarrow \sin ^{-1}(\sin 2 \alpha)+\cos ^{-1}(\cos 2 \alpha)=\frac{2 \pi}{3} \\ & \Rightarrow 2 \alpha+2 \alpha=\frac{2 \pi}{3} \\ & \Rightarrow \alpha=\frac{2 \pi}{12}=\frac{\pi}{6}\end{aligned}\)