If $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2}$, find $x$.
Step-by-Step Solution
Key Concept: When the sum of an inverse tangent and inverse sine equals $\frac{\pi}{2}$, the domain constraints force the arguments to satisfy a specific relationship.
The equation $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{2}$ holds if $x^2 + x \geq 0$ and $0 \leq x^2 + x + 1 \leq 1$. From the second constraint: $x^2 + x + 1 \leq 1$ gives $x^2 + x \leq 0$, so $x(x+1) \leq 0$. Combined with $x^2 + x \geq 0$, we get $x^2 + x = 0$, giving $x = 0$ or $x = -1$. Both values satisfy the original equation.
Correct Answer: A