Answer
$\frac{dr}{d\theta}=2\frac{1+\sin\theta}{1-\cos\theta}\frac{{(\cos\theta-1-\sin\theta)}}{({1-\cos\theta})^2}$
Work Step by Step
given $r=(\frac{1+\sin\theta}{1-\cos\theta})^2$
on differentiating both sides:
$\frac{dr}{d\theta}=\frac{d((\frac{1+\sin\theta}{1-\cos\theta})^2)}{d\theta}$
$\frac{dr}{d\theta}=2\frac{1+\sin\theta}{1-\cos\theta}(\frac{({1-\cos\theta})\frac{d({1+\sin\theta})}{d\theta}-{(1+\sin\theta)\frac{d({1-\cos\theta})}{d\theta}}}{({1-\cos\theta})^2})$
$\frac{dr}{d\theta}=2(\frac{1+\sin\theta}{1-\cos\theta})\frac{({1-\cos\theta}){\cos\theta}-{(1+\sin\theta){({\sin\theta})}})}{({1-\cos\theta})^2}$
$\frac{dr}{d\theta}=2\frac{1+\sin\theta}{1-\cos\theta}\frac{{(\cos\theta-1-\sin\theta)}}{({1-\cos\theta})^2}$
