Answer
The Derivative is:
$\frac{dr}{d\theta}=\sec^2\theta-\csc^2\theta$
Work Step by Step
$r=\sec\theta\csc\theta$
Applying Derivative rules:
$y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$
$\frac{dr}{d\theta}=(\csc\theta)\cdot\frac{d}{d\theta}(\sec\theta)+(\sec\theta)\cdot\frac{d}{d\theta}(\csc\theta)$
$\frac{dr}{d\theta}=(\csc\theta)(\sec\theta\tan\theta)+(\sec\theta)(-\csc\theta\cot\theta)$
$\frac{dr}{d\theta}=\sec\theta\csc\theta\tan\theta-\sec\theta\csc\theta\cot\theta$
$\frac{dr}{d\theta}=\sec\theta\csc\theta(\tan\theta-\cot\theta)$
Applying trigonometric identities
$\frac{dr}{d\theta}=\frac{1}{\cos\theta}\frac{1}{\sin\theta}(\frac{sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta})$
$\frac{dr}{d\theta}=\frac{1}{\cos\theta\sin\theta}\cdot\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta\sin\theta}\cdot\frac{\cos\theta}{\sin\theta}$
$\frac{dr}{d\theta}=\frac{1}{\cos^2\theta}-\frac{1}{\sin^2\theta}$
$\frac{dr}{d\theta}=\sec^2\theta-\csc^2\theta$
