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

Inverse Trigonometric Functions

Solution:

$\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$
$x ( x +1) \geq 0$.....(1)
image
Also $0 \leq x ^2+ x +1 \leq 1$
but $x^2+x+1$ is always $\geq 0$
$\therefore x^2+x+1 \leq 1 $
$x(x-1) \leq 0$.....(2)
$\operatorname{sum}(1) \text { and }(2) x(x+1)=0 $
$\Rightarrow x=0 \text { or } x=-1$