Question

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

Answer 1

${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\pi /2\to \left(i\right)$

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

$\sqrt{x^2+x}=\tan \theta$

$\sec \theta =\sqrt{1+{\tan }^2\theta }$

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

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

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

${\cos }^{-1}\alpha +{\sin }^{-1}\alpha =\frac{\pi }{2}$ is possible when,

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

$x^2+x+1=1$

$x^2+x=0$

$x(x+1)=0$

$x=0$, $x=1$