Find the real solution of: tan^-1 root(x(x+1)) + sin^-1root(x²+x+1) = pi/2.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ {\tan }^{-1}\left(\sqrt{\mathrm{x}\left(\mathrm{x}+1\right)}\right)+{\sin }^{-1}\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\tan }^{-1}\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}}\right)+{\sin }^{-1}\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{2} ;\left(\mathrm{i}\right) \\ \mathrm{Let} {\tan }^{-1}\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}}\right)=\mathrm{\theta } \\ \Rightarrow \sqrt{{\mathrm{x}}^2+\mathrm{x}}=\tan \left(\mathrm{\theta }\right) \\ \mathrm{sec\theta }=\sqrt{1+{\tan }^2\mathrm{\theta }}=\sqrt{1+{\mathrm{x}}^2+\mathrm{x}} \\ \Rightarrow \mathrm{sec\theta }=\sqrt{1+{\mathrm{x}}^2+\mathrm{x}} \\ \Rightarrow \cos \left(\mathrm{\theta }\right)=\frac{1}{\sqrt{1+{\mathrm{x}}^2+\mathrm{x}}} \\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2+\mathrm{x}}}\right) \\ \mathrm{Hence} \left(\mathrm{i}\right) \mathrm{becomes}: \\ {\cos }^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2+\mathrm{x}}}\right)+{\sin }^{-1}\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{2} ;\left(\mathrm{ii}\right) \\ \mathrm{But} \mathrm{we} \mathrm{know} \mathrm{that} \\ {\cos }^{-1}\left(\mathrm{\alpha }\right)+{\sin }^{-1}\left(\mathrm{\alpha }\right)=\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence} \left(\mathrm{ii}\right) \mathrm{is} \mathrm{possible} \mathrm{when}: \\ \frac{1}{\sqrt{1+{\mathrm{x}}^2+\mathrm{x}}}=\sqrt{{\mathrm{x}}^2+\mathrm{x}+1} \\ \Rightarrow {\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)}^2=1 \\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}+1=1 \\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}=1 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}+1\right)=0 \\ \Rightarrow \mathbf{x}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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