If \(\tan ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}\), then the value of \(x\) is

\( \tan ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2} \)
\( \Rightarrow \tan ^{-1} \sqrt{x^2+x}=\frac{\pi}{2}-\sin ^{-1} \sqrt{x^2+x+1}=\cos\)\(^{-1} \sqrt{x^2+x+1} \)
\( \Rightarrow \tan ^{-1} \sqrt{x^2+x}=\tan ^{-1} \frac{\sqrt{-\left(x^2+x\right)}}{\sqrt{x^2+x+1}} ; \sqrt{-\left(x^2+x\right)} \)
\( \Rightarrow x^2+x=\frac{-\left(x^2+x\right)}{x^2+x+1} \)
\( \Rightarrow\left(x^2+x\right)\left(x^2+x+1\right)=-\left(x^2+x\right)\) \(\left[\text { let } x^2+x=y\right] \)
\( \Rightarrow y(y+1)=-y \)
\( \Rightarrow y^2+2 y=0 \)
\( \Rightarrow y(y+2)=0 \)
\( \Rightarrow y=0 \text { or } y=-2 \)
\( \Rightarrow x^2+x=0 \text { or } x^2+x=-2 \)
\( \Rightarrow x(x+1)=0 \text { or } x^2+x+2=0 \)
\( \Rightarrow(x=0,-1) \text { or } \text { (No solution) }\)