Prove the identity
\(\dfrac{\tan\theta}{1+\sec\theta} + \dfrac{1+\sec\theta}{\tan\theta} \equiv \dfrac{2}{\sin\theta}\)
Solution
\(\dfrac{\tan^2\theta + (1+\sec\theta)^2}{\tan\theta(1+\sec\theta)}\)
\(= \dfrac{\tan^2\theta + 1 + 2\sec\theta + \sec^2\theta}{\tan\theta(1+\sec\theta)}\)
\(= \dfrac{\sec^2\theta + 2\sec\theta + \sec^2\theta}{\tan\theta(1+\sec\theta)}\)
\(= \dfrac{2\sec\theta(1+\sec\theta)}{\tan\theta(1+\sec\theta)}\)
\(= \dfrac{2\sec\theta}{\tan\theta}\)
\(= 2 \cdot \dfrac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}\)
\(= \dfrac{2}{\sin\theta}\)
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