Prove \( \cos ^{2} \theta \) \( \frac{\sin x+\cos x}{1+\tan x}=\cos x \)

Let's start with the equation: \[ \frac{\sin x + \cos x}{1 + \tan x} = \cos x \] To simplify this, we can rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\): \[ 1 + \tan x = 1 + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x} \] Now substitute this back into the equation: \[ \frac{\sin x + \cos x}{\frac{\cos x + \sin x}{\cos x}} = \cos x \] By simplifying the left side, we multiply by the reciprocal: \[ (\sin x + \cos x) \cdot \frac{\cos x}{\cos x + \sin x} = \cos x \] If we assume \(\sin x + \cos x \neq 0\), we can cancel terms on both sides, giving us: \[ \cos x = \cos x \] This equation holds true for all \(x\) where \(\sin x + \cos x \neq 0\). Thus, we have verified the identity \(\frac{\sin x + \cos x}{1 + \tan x} = \cos x\) is valid.