Question

Find the real solution of ${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$.

Answer 1

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

$\tan y=\sqrt{x(x+1)}$

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

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

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

We know that

${\sin }^{-1}y+{\cos }^{-1}y=\frac{\pi }{2}$

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

$x^2+x+1=1$

$x^2+x=0$

$x=0,-1$