# Prove the Following. Tan θ S E C θ + 1 = S E C θ − 1 Tan θ - Geometry

#### Question

Prove the following.
$\frac{\tan\theta}{sec\theta + 1} = \frac{sec\theta - 1}{\tan\theta}$

#### Solution

$\frac{\tan\theta}{\sec\theta + 1}$
$= \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}$
$= \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}$
$= \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \left( 1 + \tan^2 \theta = \sec^2 \theta \right)$
$= \frac{\sec\theta - 1}{\tan\theta}$

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