If $$\dfrac{(\sin \theta - \cosec \theta )(\cos \theta - \sec \theta)}{\tan^2 \theta - \sin^2 \theta} = r^3,  then  r =$$

$$\dfrac{(\sin \theta - \cosec \theta )(\cos \theta - \sec \theta)}{\tan^2 \theta - \sin^2 \theta} = r^3$$

we know that $$\cosec\theta=\dfrac{1}{\sin\theta}$$ & $$\sec\theta=\dfrac{1}{\cos\theta}$$ and $$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$$

$$\dfrac{(\sin \theta - \dfrac{1}{\sin\theta} )(\cos \theta -\dfrac{1}{\cos\theta})}{\dfrac{\sin^2\theta}{\cos^2\theta}- \sin^2 \theta} = r^3$$

$$\dfrac{\dfrac{(\sin^2 \theta-1}{\sin\theta} )(\dfrac{\cos^2 \theta -1}{\cos\theta})}{\dfrac{\sin^2\theta+\sin^2\theta\cos^2\theta}{\cos^2\theta}} = r^3$$

from identity $$\sin^2\theta+\cos^2\theta=1$$

$$\cos^2\theta=1-\sin^2\theta$$

$$\sin^2\theta=1-\cos^2\theta$$

$$\dfrac{\dfrac{(\cos^2 \theta}{\sin\theta} )(\dfrac{\sin^2\theta}{\cos\theta})}{\dfrac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}} = r^3$$

$$\dfrac{\dfrac{\cos \theta}{\sin\theta}}{\dfrac{\sin^4\theta}{\cos^2\theta}} = r^3$$

$$\dfrac{\cos^3 \theta}{\sin^3\theta} = r^3$$

$$r=\cot \theta$$