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Question
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Solution
\[\frac{\tan\theta}{\sec\theta + 1}\]
\[ = \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}\]
\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}\]
\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \] ...(1 + tan2θ = sec2θ)
\[ = \frac{\sec\theta - 1}{\tan\theta}\]
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