Question

Prove the following.

\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]

Answer 1

\[\frac{\tan\theta}{\sec\theta + 1}\]

\[ = \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}\]

\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}\]

\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \]       ...(1 + tan²θ = sec²θ)

\[ = \frac{\sec\theta - 1}{\tan\theta}\]