IIT-JEE 1999Question 1
The number of real solutions of $\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^{2}+x+1}=\pi/2$ is
Inverse Trigonometric Functions
Solution
Need $x(x+1)\ge0$ and $x^2+x+1\le1$, i.e. $x(x+1)\le0$, so $x(x+1)=0$: $x=0$ or $-1$. Both give $\tan^{-1}0+\sin^{-1}1=\frac\pi2$. Two solutions.