Answer:
We have, \[{{\tan }^{-1}}\sqrt{{{x}^{2}}+x}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2}\] This equation holds, if \[{{x}^{2}}+x\ge 0\] and \[0\le {{x}^{2}}+x+1\le 1\] Now, \[{{x}^{2}}+x\ge 0\] and \[0\le {{x}^{2}}+x+1\le 1\] \[\Rightarrow \] \[{{x}^{2}}+x\ge 0\] and \[{{x}^{2}}+x+1\le 1\] \[[\because {{x}^{2}}+x+1>0\,\,\text{for}\,\,\text{all}\,\,x]\] \[\Rightarrow \] \[{{x}^{2}}+x\ge 0\] and \[{{x}^{2}}+x\le 0\] \[\Rightarrow \] \[{{x}^{2}}+x=0\] \[\Rightarrow \] \[x(x+1)=0\] \[\Rightarrow \] \[x=0,\,\,-\,1\] Hence, \[x=0,\,\,-\,1\] are the solution of the given equation.
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