Question

solve: tan^-1 (under root x² +x) +sin^-1 (under root x²+x+1) = pie/2 ??

Answer 1

$$\begin{gathered} {\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\tan }^{-1}\sqrt{x^2+x}=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\sqrt{x^2+x+1} \\ \Rightarrow {\tan }^{-1}\sqrt{x^2+x}={\cos }^{-1}\sqrt{x^2+x+1} \end{gathered}$$

$$\begin{gathered} \Rightarrow \sqrt{x^2+x}=\frac{\sqrt{1-\left(x^2+x+1\right)}}{\sqrt{x^2+x+1}}\left[\because \tan \left({\cos }^{-1}x\right)=\frac{\sqrt{1-x^2}}{x}\right] \\ \Rightarrow \left(x^2+x\right)=\frac{1-\left(x^2+x+1\right)}{{\left(x^2+x+1\right)}^2} \\ \Rightarrow \left(x^2+x\right)\left(x^2+x+1\right)=-\left(x^2+x\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow x^2+x+2=0 \\ \Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times 2}}{2\times 1} \\ \Rightarrow x=\frac{-1+i\sqrt{7}}{2},or,x=\frac{-1-i\sqrt{7}}{2} \end{gathered}$$