Answer
$1 - \sin x $
Work Step by Step
Given expression is-
$ \frac{\cos x}{\sec x + \tan x} $
= $ \frac{\cos x}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} $
(Writing in terms of sin and cos)
= $ \frac{\cos x}{\frac{1 + \sin x}{\cos x} } $
= $ \cos x . \frac {\cos x}{1 + \sin x } $
= $ \frac {\cos^{2} x}{1 + \sin x } $
= $ \frac {1 - \sin^{2} x}{1 + \sin x } $
( From first Pythagorean identity, $\cos^{2} x$ = $1 - \sin^{2} x$)
= $ \frac {(1 - \sin x) (1 + \sin x)}{1 + \sin x } $
{Recall $a^{2} - b^{2}$ = (a-b)(a+b)}
= $1 - \sin x $
