Explanation

(d)
Given pair of line is
$x^{2}\sin^{2}\alpha + y^{2}\sin^{2}\alpha = x^{2}\cos^{2}\theta + y^{2}\sin^{2}\theta - 2xy\sin{\theta\cos\theta}$
$\Rightarrow \ x^{2}\left( \sin^{2}\alpha - \cos^{2}\theta \right){+ y}^{2}\left( \sin^{2}\alpha - \sin^{2}\theta \right) + 2(\sin{\theta\cos{\theta)xy = 0}}$
On comparing with $ax^{2} + {by}^{2} + 2hxy = 0$
We get, $a = \sin^{2}\alpha - \cos^{2}\theta,$
$b = \sin^{2}\alpha - \sin^{2}\theta\ and\ h = \sin{\theta\cos\theta}$
Let $\theta$ be the angle between the pair of lines.
$\therefore\tan{\theta =}\left| \frac{2\sqrt{\sin^{2}\theta\cos^{2}\theta - \left( \sin^{2}\alpha - \cos^{2}\theta \right) \times \left( \sin^{2}\alpha - \sin^{2}\theta \right)}}{\sin^{2}\alpha - \cos^{2}\theta + \sin^{2}\alpha - \sin^{2}\theta} \right|$
$= \left| \frac{2\sqrt{\sin^{2}\theta\cos^{2}\theta - \left( \sin^{2}\alpha \right)^{2} + \sin^{2}\alpha{\ sin}^{2}\theta + \sin^{2}\alpha\ \cos^{2}\theta - \sin^{2}\theta\ \cos^{2}\theta}}{- \left( 1 - 2\sin^{2}\alpha \right)} \right|$
$= \left| \frac{2\sqrt{\sin^{2}\alpha\left( \sin^{2}\theta + \cos^{2}\theta \right) - \left( \sin^{2}\alpha \right)^{2}}}{- \cos{2\ \alpha}} \right|$
$= \left| \frac{2\sqrt{\sin^{2}\alpha\left( 1 - \sin^{2}\alpha \right)}}{- \cos{2\alpha}} \right|$
$\Rightarrow \ \tan{\theta = \left| \frac{\sin{2\alpha}}{\cos{2\alpha}} \right| = \tan{2\alpha}}$
$\Rightarrow \ \theta = 2\alpha$