Question

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

Answer 1

Given :
${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$
Consider ${\tan }^{-1}\sqrt{x(x+1)}$
Domain of ${\tan }^{-1}x$ is $R$
$\Rightarrow x(x+1)\ge 0$
[As $\sqrt{x}$ is defined for $x\ge 0$ ]
$\Rightarrow x^2+x\ge 0$
$\Rightarrow x^2+x+1\ge 1\dots \left(1\right)$
Consider ${\sin }^{-1}\sqrt{x^2+x+1}$
$\because$ Domain of ${\sin }^{-1}x$ is $[-1,1]$
$\Rightarrow -1\le \sqrt{x^2+x+1}\le 1$
$\Rightarrow 0\le \sqrt{x^2+x+1}\le 1[\because \sqrt{x}\ge 0]$
$\Rightarrow 0\le \sqrt{x^2+x+1\le 1}\dots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$\Rightarrow x^2+x+1=1$
$\Rightarrow x^2+x=0$
$\Rightarrow x(x+1)=0$
$\Rightarrow x=0,-1$
$\Rightarrow x=0,-1$
$\Rightarrow {\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}$
$={\tan }^{-1}\sqrt{0}+{\sin }^{-1}\sqrt{1}$
$\Rightarrow {\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=0+\frac{\pi }{2}$
$\Rightarrow {\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$
$\therefore x=0$ and $x=-1$