If $\frac{1}{2} \leq x \leq 1$, then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & 	herefore \cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3}}{2} \sqrt{1-x^2}ight) \ & =\cos ^{-1} x+\cos ^{-1}\left(x 	imes \frac{1}{2}+\frac{\sqrt{3}}{2} 	imes \sqrt{1-x^2}ight) \ & =\cos ^{-1} x+\cos ^{-1}\left(\frac{1}{2}ight)-\cos ^{-1} x \ & =\cos ^{-1}\left(\frac{1}{2}ight)=\frac{\pi}{3}\end{aligned}$

If $\frac{1}{2} \leq x \leq 1$, then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3…

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