Question

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

Answer 1

We have ${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+x}=\frac{\pi }{2}......\left(i\right)$.

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

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

$\Rightarrow \tan \theta =\frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}}[\because \tan \theta =\frac{\sin \theta }{\cos \theta }]$

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

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

On putting the value of $\theta$ in equ (i) we get

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

We know that, ${\tan }^{-1}x+{\tan }^{-1}\left(\frac{x+y}{1-xy}\right),xy<1$

${\tan }^{-1}\left[\frac{\sqrt{x(x+1)}+\sqrt{\frac{x^2+x+1}{-x^2-x}}}{1-\sqrt{x(x+1)}\sqrt{\frac{x^2+x+1}{-x^2-x}}}\right]=\frac{\pi }{2}$

${\tan }^{-1}\left[\frac{\sqrt{x^2+x}+\sqrt{\frac{x^2+x+1}{-1(x^2+x)}}}{1-\sqrt{x^2+x}\sqrt{\frac{x^2+x+1}{-1(x^2+x)}}}\right]=\frac{\pi }{2}$

$\Rightarrow \frac{x^2+x+\sqrt{-(x^2+x+1)}}{[1-\sqrt{-(x^2+x+1)}\sqrt{(x^2+x)}]}=\tan \frac{\pi }{2}=\frac{1}{0}$

$\Rightarrow [1-\sqrt{-(x^2+x+1)}]\sqrt{(x^2+x)}=0\Rightarrow =(x^2+x+1)=1$ or $x^2+x=0$

$\Rightarrow -x^2-x-1=1$ or $x(x+1)=0$

$\Rightarrow x^2+x+2=0$ or $x(x+1)=0$

$\therefore x=\frac{1\pm \sqrt{1-4\times 2}}{2}$

$\Rightarrow x=0$ or $x=-1$

For real solution, we have $x=0,1$.