Question

Solve the equation ${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$

Answer 1

${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$

${\tan }^{-1}\sqrt{x^2+x}=\frac{\pi }{2}-{\sin }^{-1}\sqrt{x^2+x+1}$

${\tan }^{-1}\sqrt{x^2+x}={\cos }^{-1}\sqrt{x^2+x+1}$

Apply ${\tan }^{-1}y={\cos }^{-1}\frac{1}{\sqrt{1+y^2}}$ on R.H.S.

${\cos }^{-1}\frac{1}{\sqrt{1+(\sqrt{x^2+x}{)}^2}}={\cos }^{-1}\sqrt{x^2+x+1}$

${\cos }^{-1}\frac{1}{\sqrt{x^2+x+1}}={\cos }^{-1}\sqrt{x^2+x+1}$

$\Rightarrow \frac{1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}$

$x^2+x+1=1$

$x^2+x=0$

$x(x+1)=0$

$x=0,-1$