The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$        is

Solution :
Clearly, $$x\left( {x + 1} \right) \geqslant 0$$   and $${x^2} + x + 1 \leqslant 1.$$    Together they imply $$x\left( {x + 1} \right) = 0.$$
∴ $$x = 0, -1.$$   When $$x = 0,$$  L.H.S. $$ = {\tan^{ - 1}}0 + {\sin ^{ - 1}}1 = \frac{\pi }{2}.$$
When $$x = -1,$$   L.H.S. $$ = {\tan^{ - 1}}0 + {\sin ^{ - 1}}\sqrt {1 - 1 + 1} = 0 + {\sin ^{ - 1}}1 = \frac{\pi }{2}.$$