Question

Prove the trigonometric identity $\frac{\tan \theta }{1+{\tan }^2\theta }+\frac{\cot \theta }{1+{\cot }^2\theta }=2\sin \theta \cos \theta$. Using the identities $1+{\tan }^2\theta ={\sec }^2\theta$ and $1+{\cot }^2\theta ={\csc }^2\theta$, the LHS becomes $\frac{\sin \theta }{\cos \theta }\cdot {\cos }^2\theta +\frac{\cos \theta }{\sin \theta }\cdot {\sin }^2\theta =\sin \theta \cos \theta +\cos \theta \sin \theta =2\sin \theta \cos \theta$.