Explanation

(d)
Given that, $x = a\sin{\theta\ and\ y = b\cos\theta}$
On differentiating w.r.t. $\theta$, we get
$\frac{dx}{d\theta} = a\cos{\theta\ and\ }\frac{dy}{d\theta} = - b\sin\theta$
$\Rightarrow \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{- b}{a}\tan\theta$
Again differentiating w.r.t. $\theta$, we get
$\frac{d^{2}y}{dx^{2}} = \frac{- b}{a}\sec^{2}\theta\frac{d\theta}{dx}$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = - \frac{b}{a}\sec^{2}\theta\frac{1}{a\cos\theta} = - \frac{b}{a^{2}}sec^{3}\ \theta$