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

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Aicharya · Accepted Answer

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 $\begin{aligned} & \sin ^{-1}\left(2 \cos \alpha \sqrt{1-\cos ^2 \alpha}\right)+\sec ^{-1}\left(\frac{1}{2 \cos ^2 \alpha-1}\right)=\frac{2 \pi}{3} \\ & \Rightarrow \sin ^{-1}(2 \cos \alpha \sin \alpha)+\sec ^{-1}\left(\frac{1}{\cos 2 \alpha}\right)=\frac{2 \pi}{3} \end{aligned}$ $\begin{aligned} & \Rightarrow \sin ^{-1}(\sin 2 \alpha)+\cos ^{-1}(\cos 2 \alpha)=\frac{2 \pi}{3} \\ & \Rightarrow 2 \alpha+2 \alpha=\frac{2 \pi}{3} \\ & \Rightarrow \alpha=\frac{2 \pi}{12}=\frac{\pi}{6}\end{aligned}$

If $\cos ^{-1} x=\alpha(0 \lt x \lt 1)$ and $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+\sec…

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