Answer
As left side transforms into right side, hence given identity-
$\sec x + \tan x $ = $ \frac{1}{\sec x - \tan x}$ is true.
Work Step by Step
Given identity is-
$\sec x + \tan x $ = $ \frac{1}{\sec x - \tan x}$
Taking L.S.
$\sec x + \tan x $
= $(\sec x + \tan x) . \frac{(\sec x - \tan x)}{(\sec x - \tan x)} $
{Multiplying and dividing by , $(\sec x - \tan x)$}
= $\frac{(\sec x + \tan x)(\sec x - \tan x)}{ (\sec x - \tan x)} $
= $\frac{ \sec^{2} x - \tan^{2} x} {\sec x - \tan x}$
{Recall $(a+b)(a-b) $ = $a^{2} - b^{2}$ }
=$ \frac{1}{\sec x - \tan x}$
( From second Pythagorean identity, $\sec^{2} x - \tan^{2} x$ = $1$)
= R.S.
As left side transforms into right side, hence given identity-
$\sec x + \tan x $ = $ \frac{1}{\sec x - \tan x}$ is true.
