Question

Prove

$tan\Theta +sec\theta =\frac{1}{sec\Theta -tan\Theta }$

Answer 1

$\tan \theta +\sec \theta =\frac{1}{\sec \theta -\tan \theta }$

Solving LHS$=\tan \theta +\sec \theta$ (on rationalising)

$\Rightarrow \frac{(\tan \theta +\sec \theta )(\tan \theta -\sec \theta )}{\tan \theta -\sec \theta }=\frac{{\tan }^2\theta -{\sec }^2\theta }{\tan \theta -\sec \theta }$

We know $1+{\tan }^2\theta ={\sec }^2\theta$

$\Rightarrow {\tan }^2\theta -{\sec }^2\theta =-1$

Then

$\Rightarrow \frac{-1}{\tan \theta -\sec \theta }$

$\Rightarrow \frac{1}{\sec \theta -\tan \theta }=$RHS.