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

$x = 0, -1$

The correct answer is Option (3) → $x = 0, -1$ ##

We have, $\tan^{-1} \sqrt{x(x + 1)} + \sin^{-1} \sqrt{x^2 + x + 1} = \frac{\pi}{2} \dots \text{(i)}$

Let $\tan^{-1} \sqrt{x(x + 1)} = \alpha \Rightarrow \sqrt{x^2 + x} = \tan \alpha$

$\Rightarrow \tan^2 \alpha = x^2 + x \dots \text{(ii)}$

Again let $\sin^{-1} \sqrt{x^2 + x + 1} = \beta \Rightarrow \sqrt{x^2 + x + 1} = \sin \beta$

$\Rightarrow \sin^2 \beta = x^2 + x + 1 \dots \text{(iii)}$

From Eq. (i), we get

$\alpha + \beta = \frac{\pi}{2} \Rightarrow \beta = \frac{\pi}{2} - \alpha \dots \text{(iv)}$$

From Eqs. (ii) and (iii), we get

$\sin^2 \beta = \tan^2 \alpha + 1 \Rightarrow \sin^2 \beta = \sec^2 \alpha \quad [∵1 + \tan^2 \theta = \sec^2 \theta]$

$\Rightarrow \sin^2 \beta \cos^2 \alpha = 1 \Rightarrow \sin^2 \left( \frac{\pi}{2} - \alpha \right) \cos^2 \alpha = 1 \quad [\text{from Eq. (iv)}]$

$\Rightarrow \cos^2 \alpha \cos^2 \alpha = 1 \Rightarrow \cos^4 \alpha = \cos 0 \Rightarrow \alpha = 0$

On putting value of $\alpha$ in Eq. (ii), we get:

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