Explanation

(d)
Given, $\tan A$ and $\tan B$ are the roots of the equation
$abx^{2} - c^{2}x + ab = 0$
$\therefore\ \tan{A + \tan{B = \frac{c^{2}}{ab}},\tan A\tan{B = 1}}$
Now, $\tan{(A + B) = \frac{\tan{A + \tan B}}{1 - \tan A\tan B} = \frac{\frac{c^{2}}{ab}}{1 - 1}} = \infty$
$\Rightarrow \ \ A + B = \frac{\pi}{2}\ \ \Rightarrow \ \ C = \frac{\pi}{2}$
$\therefore\ \sin^{2}A + \sin^{2}{B + \sin^{2}C}$
$= \sin^{2}{\left( \frac{\pi}{2} - B \right) + \sin^{2}{B + \sin^{2}\frac{\pi}{2}}}$
$= \cos^{2}{B + \sin^{2}{B + 1 = 2}}$