\( \frac { \sin x + \cos x } { 1 + \tan } = \cos x . \)

To solve the equation \( \frac { \sin x + \cos x } { 1 + \tan x } = \cos x \), we can start by substituting the identity \( \tan x = \frac{\sin x}{\cos x} \). This changes our equation to: \[ \frac{\sin x + \cos x}{1 + \frac{\sin x}{\cos x}} = \cos x. \] Now, multiplying the denominator by \(\cos x\) to clear the fraction gives: \[ \frac{\sin x + \cos x}{\frac{\cos x + \sin x}{\cos x}} = \cos x. \] This simplifies to: \[ \frac{(\sin x + \cos x) \cos x}{\cos x + \sin x} = \cos x, \] which can further reduce to: \[ \sin x + \cos x = \cos x + \sin x. \] Since both sides are equal, it suggests that the original equation holds true for all \( x \) where the expressions are defined. However, we must consider any restrictions. The equation \( 1 + \tan x \neq 0 \) or \( \tan x \neq -1 \) also needs to be satisfied, which happens at angles such as \( x = \frac{3\pi}{4} + k\pi \) for \( k \in \mathbb{Z} \). This shows that the equation holds for all \( x \) except those specific angles where \( \tan x = -1 \).