Question
Prove the following.
$\frac{\tan \theta}{\sec \theta+1}=\frac{\sec \theta-1}{\tan \theta}$
$\frac{\tan \theta}{\sec \theta+1}=\frac{\sec \theta-1}{\tan \theta}$
✓
Answer
Taking LHS
$=\frac{\tan \theta}{\sec \theta+1} \times \frac{\sec \theta-1}{\sec \theta-1}$
$=\frac{\tan \theta(\sec \theta-1)}{\sec ^2 \theta-1}$
$=\frac{\tan \theta\left(\sec ^2 \theta-1\right)}{\tan ^2 \theta}\left[\tan ^2 \theta=\sec ^2 \theta-1\right]$
$=\frac{\sec \theta-1}{\tan \theta}$
= RHS
Proved !
$=\frac{\tan \theta}{\sec \theta+1} \times \frac{\sec \theta-1}{\sec \theta-1}$
$=\frac{\tan \theta(\sec \theta-1)}{\sec ^2 \theta-1}$
$=\frac{\tan \theta\left(\sec ^2 \theta-1\right)}{\tan ^2 \theta}\left[\tan ^2 \theta=\sec ^2 \theta-1\right]$
$=\frac{\sec \theta-1}{\tan \theta}$
= RHS
Proved !
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