Let $$ - 1 \leqslant x \leqslant 1.$$   If $$\cos\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{2},$$    then how many value does $$\tan\left( {{{\cos }^{ - 1}}x} \right)$$   assume ?

Solution :
As given : $$\cos\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{2}$$
$$\eqalign{ & \Rightarrow {\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) \cr & \Rightarrow x = \sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2} \cr & \therefore \tan\left( {{{\cos }^{ - 1}}x} \right) = \tan \left( {{{\cos }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right) \cr & = \tan \left( { \pm \frac{\pi }{6}} \right) = \pm \frac{1}{{\sqrt 3 }} \cr} $$
Hence, $$\tan\left( {{{\cos }^{ - 1}}x} \right)$$   have two values.