If x=2 sinθ - sin2θ and y=2 cosθ - cos2θ ,θ ∈ [0,2π ], then d^2y dx^2 at θ =π is:

$$\frac{dx}{d\theta }=2\mathrm{cos}\theta -2\mathrm{cos}2\theta$$dθdx​=2cosθ−2cos2θ

$$\frac{dy}{d\theta }=-2\mathrm{sin}\theta +2\mathrm{sin}2\theta$$dθdy​=−2sinθ+2sin2θ

$$\therefore \frac{dy}{dx}=\frac{\mathrm{sin}2\theta -\mathrm{sin}\theta }{\mathrm{cos}\theta -\mathrm{cos}2\theta }$$∴dxdy​=cosθ−cos2θsin2θ−sinθ​

$$=\frac{2\mathrm{sin}\frac{\theta }{2}.\mathrm{cos}\frac{3\theta }{2}}{2\mathrm{sin}\frac{\theta }{2}.\mathrm{sin}\frac{3\theta }{2}}=\mathrm{cot}\frac{3\theta }{2}$$=2sin2θ​.sin23θ​2sin2θ​.cos23θ​​=cot23θ​

$$\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{d\theta }(\frac{dy}{dx})\frac{d\theta }{dx}=-\frac{3}{2}{\mathrm{cosec}}^{2}\frac{3\theta }{2}.\frac{d\theta }{dx}$$dx2d2y​=dθd​(dxdy​)dxdθ​=−23​cosec223θ​.dxdθ​

$$\Rightarrow \frac{{d}^{2}y}{d{x}^{2}}=\frac{-\frac{3}{2}{\mathrm{cosec}}^{2}\frac{3\theta }{2}}{2(\mathrm{cos}\theta -\mathrm{cos}2\theta )}$$⇒dx2d2y​=2(cosθ−cos2θ)−23​cosec223θ​​

$$\Rightarrow {\frac{{d}^{2}y}{d{x}^{2}}|}_{\theta =\pi }=\frac{3}{4(-1-1)}=\frac{3}{8}$$⇒dx2d2y​∣θ=π​=4(−1−1)3​=83​