Prove the identity
\(\dfrac{\cot\theta}{1+\cot^2\theta} \equiv \sin\theta\cos\theta\)
Solution
\(\dfrac{\cot\theta}{1+\cot^2\theta} = \dfrac{\cot\theta}{\csc^2\theta}\) \(\;\) [since \(1+\cot^2\theta = \csc^2\theta\)]
\(= \cot\theta \cdot \sin^2\theta\) \(\;\) [since \(\dfrac{1}{\csc^2\theta} = \sin^2\theta\)]
\(= \dfrac{\cos\theta}{\sin\theta} \cdot \sin^2\theta\)
\(= \cos\theta \sin\theta\)
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