Given,
$\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$
$\Rightarrow \quad \cos ^{-1} \frac{1}{\sqrt{\left(x^{2}+x\right)^{2}+1}}$
$=\frac{\pi}{2}-\sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}$
$\Rightarrow \cos ^{-1} \frac{1}{\sqrt{\left(x^{2}+x\right)^{2}+1}}=\cos ^{-1} \sqrt{x^{2}+x+1}$
$\Rightarrow \quad \frac{1}{\sqrt{\left(x^{2}+x\right)^{2}+1}}=\sqrt{x^{2}+x+1}$
$\Rightarrow \quad 1=\left(x^{2}+x+1\right)\left[\left(x^{2}+x\right)^{2}+1\right]$
$\Rightarrow \quad\left(\mathrm{x}^{2}+\mathrm{x}\right)^{3}+\left(\mathrm{x}^{2}+\mathrm{x}\right)^{2}+\left(\mathrm{x}^{2}+\mathrm{x}\right)+1=1$
$\Rightarrow \quad\left(x^{2}+x\right)\left[\left(x^{2}+x\right)^{2}+\left(x^{2}+x\right)+1\right]=0$
$\Rightarrow \quad x^{2}+x=0$
$$
\Rightarrow \quad \mathrm{x}=0,-1
$$
Hence, both values of $\mathrm{x}$ satisfies the given equation.