Solution

Given:

\[ \tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2} \tag{i} \]

Let \[ \sin^{-1}\sqrt{x^2 + x + 1} = \theta \]

\[ \Rightarrow \sin\theta = \sqrt{x^2 + x + 1} \]

\[ \Rightarrow \tan\theta = \frac{\sqrt{x^2 + x + 1}}{\sqrt{1 – (x^2 + x + 1)}} = \frac{\sqrt{x^2 + x + 1}}{\sqrt{-x^2 – x}} \]

Substituting value of \(\theta\) in equation (i), we get:

\[ \tan^{-1}\sqrt{x(x+1)} + \tan^{-1}\left( \frac{\sqrt{x^2 + x + 1}}{\sqrt{-x^2 – x}} \right) = \frac{\pi}{2} \]

Using identity: \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad xy < 1 \]

\[ \tan^{-1}\left[ \frac{ \sqrt{x(x+1)} + \sqrt{\frac{x^2 + x + 1}{-x^2 – x}} }{ 1 – \sqrt{x(x+1)}\cdot \sqrt{\frac{x^2 + x + 1}{-x^2 – x}} } \right] = \frac{\pi}{2} \]

\[ \Rightarrow \left[1 – \sqrt{-(x^2 + x + 1)}\right]\sqrt{x^2 + x} = 0 \]

\[ \Rightarrow x^2 + x = 0 \quad \text{or} \quad x^2 + x + 2 = 0 \]

\[ \Rightarrow x(x+1) = 0 \]

\[ \boxed{x = 0 \text{ or } x = -1} \]

Hence, the real solutions are \(x = 0\) and \(x = -1\).