Given that,

$${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$

We know,

$$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$$

$$\therefore$$ $$\,\,\,$$ $$ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$$

$$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$$

$$ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$$

$$ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$$

$$\therefore$$ $$\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$$

Given that,

$${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( {\cos x} \right) = x\,\,\,\,...\left( 1 \right)$$

We know,

$${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$$

$$\therefore$$ $$\,\,\,$$ $${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$$

Adding $$(1)$$ and $$(2),$$ we get,

$$2{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}$$

$$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$$

$$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$$

$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$$

$$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$$

Squaring both sides we get,

$$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$$

$$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$$

$$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$$

Applying compounds and dividendo rule,

$$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$$

$$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$$