$x = 2\sin \theta - \sin 2\theta$ $\Rightarrow$ ${{dx} \over {d\theta }}$ = $2\cos \theta - 2\cos 2\theta$ $y = 2\cos \theta - \cos 2\theta$ $\Rightarrow$ ${{dy} \over {d\theta }}$ = –2sin$\theta$ + 2sin2$\theta$ ${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$ = ${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$ This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression: $\frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$ This is the rate of change of y with respect to x, as a function of θ. Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$, yielding : $\frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$ Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes : $\frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$ Finally, we evaluate this expression at $\theta = \pi$, yielding : $\frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$ Therefore, the correct answer is (A) $\frac{3}{8}$. Alternate Method : First, let's find the derivatives of x and y with respect to θ : 1) $\frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta$ 2) $\frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta$ We know that $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$ So, we substitute 1) and 2) into this equation : We have $\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$ For simplification, let's denote the numerator as $N = \sin 2\theta - \sin \theta$ and the denominator as $D = \cos \theta - \cos 2\theta$. We have to compute $\frac{d}{d\theta}(\frac{dy}{dx})$ which is $\frac{d}{d\theta}(\frac{N}{D})$. We can use the quotient rule for differentiation, which states that if we have a function of the form $\frac{u}{v}$, then its derivative is given by $\frac{vu' - uv'}{v^2}$. So here, $N' = 2\cos 2\theta - \cos \theta$ and $D' = -\sin \theta + 2\sin 2\theta$. Applying the quotient rule : $\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$ Substituting the expressions for $N'$, $D'$, $N$, and $D$ we get : $\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$ Now $\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}$ = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$$$ \times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)} $\begin{aligned} \left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\ & =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8} \end{aligned}$