#### Question

Prove that:

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

#### Solution

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

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

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

\[ = \sec\theta + \tan\theta \left( 1 + \tan^2 \theta = \sec^2 \theta \right)\]

Is there an error in this question or solution?

Solution Prave That: 1 Sec θ − Tan θ = Sec θ + Tan θ Concept: Application of Trigonometry.