Browse Questions

# The number of real solutions of $\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\large\frac{\pi}{2}$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;infinite$

$\tan^{-1}\sqrt{x(x+1)}=\large\frac{\pi}{2}$$-\sin^{-1}\sqrt{(x^2+x+1)} \Rightarrow \tan^{-1}\sqrt{x(x+1)}=\cos^{-1}\sqrt{x^2+x+1} \Rightarrow \cos^{-1}\large\frac{1}{\sqrt{x^2+x+1}}$$=\cos^{-1}\sqrt{x^2+x+1}$
$\Rightarrow x^2+x+1=1$
$\Rightarrow x(x+1)=0$
$x=0,-1$
Hence (c) is the correct answer.