$cos^{-1}\begin{Bmatrix}\frac{1}{2}x^2 + \sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}\end{Bmatrix}= cos^{-1}\frac{x}{2}-cos^{-1}x $ holds for
Answer & explanation
Correct answer: option 3
We observe that
$\frac{x^2}{2}+\sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}$ is positive and defined for all x ∈ [-1, 1]
$∴cos^{-1}\begin{Bmatrix}\frac{x^2}{2} + \sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}\end{Bmatrix}∈\left[0, \frac{\pi}{2}\right] $
Now, RHS of the given relation is defined for
$\left|\frac{x}{2}\right|$ and $|x| ≤ 1 $ i.e. for x ∈ [-1,1].
Also, LHS ≥ 0
$⇒ cos^{-1}\frac{x}{2}-cos^{-1} x ≥ 0 ⇒ cos^{-1}\frac{x}{2} ≥cos^{-1}x ⇒x ∈[0, 1]$