Question
Prove that:
$\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta$
$\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta$
✓
Answer
Taking LHS
$\begin{array}{l}
=\frac{1}{\sec \theta-\tan \theta} \\
=\frac{1}{\sec \theta-\tan \theta} \times \frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta} \\
=\frac{\sec \theta+\tan \theta}{\sec ^2 \theta-\tan ^2 \theta} \\
=\sec \theta+\tan \theta\left[A s, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right] \\
=\text { RHS }
\end{array}$
Proved!
$\begin{array}{l}
=\frac{1}{\sec \theta-\tan \theta} \\
=\frac{1}{\sec \theta-\tan \theta} \times \frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta} \\
=\frac{\sec \theta+\tan \theta}{\sec ^2 \theta-\tan ^2 \theta} \\
=\sec \theta+\tan \theta\left[A s, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right] \\
=\text { RHS }
\end{array}$
Proved!
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