$${\cos ^{ - 1}}\left\{ {\frac{1}{2}{x^2} + \sqrt {1 - {x^2}} \cdot \sqrt {1 - \frac{{{x^2}}}{4}} } \right\} = {\cos ^{ - 1}}\frac{x}{2} - {\cos ^{ - 1}}x$$           holds for

Solution :
Clearly, $$0 \leqslant {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \leqslant \frac{\pi }{2}$$    because $${\cos ^{ - 1}}x$$  is in the first quadrant when $$x$$ is positive.
$$\therefore \,\,{\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \geqslant 0.\,{\text{So,}}\,{\cos ^{ - 1}}\frac{x}{2} \geqslant {\cos ^{ - 1}}x.$$
Also $$\left| {\frac{x}{2}} \right| \leqslant 1,\left| x \right| \leqslant 1.$$   This means $$\left| x \right| \leqslant 1.$$
In $$0 \leqslant x \leqslant 1,$$   clearly $${\cos^{ - 1}}\frac{x}{2} \geqslant {\cos ^{ - 1}}x$$    because $${\cos ^{ - 1}}\alpha $$  increases as $$\alpha $$ decreases.