Answer: D
\(\cfrac{1}{\tan\theta+\cfrac{1}{\tan\theta}} = \cfrac{1}{\cfrac{\sin\theta}{\cos\theta}+\cfrac{\cos\theta}{\sin\theta}} = \cfrac{1}{\cfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}} = \sin\theta\cos\theta\)
\(\cfrac{1}{\tan\theta+\cfrac{1}{\tan\theta}} = \cfrac{1}{\cfrac{\sin\theta}{\cos\theta}+\cfrac{\cos\theta}{\sin\theta}} = \cfrac{1}{\cfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}} = \sin\theta\cos\theta\)