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

Ask by Stuart Mathis. in South Africa
Mar 09,2025

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\[ \frac{\sin x + \cos x}{1 + \tan x} = \cos x \]

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We start with the expression \[ \frac{\sin x+\cos x}{1+\tan x}. \] **Step 1.** Recall that \[ \tan x = \frac{\sin x}{\cos x}. \] **Step 2.** Substitute \(\tan x\) into the denominator: \[ \frac{\sin x+\cos x}{1+\frac{\sin x}{\cos x}}. \] **Step 3.** Combine the terms in the denominator by writing \(1\) as \(\frac{\cos x}{\cos x}\): \[ \frac{\sin x+\cos x}{\frac{\cos x+\sin x}{\cos x}}. \] **Step 4.** Rewrite the expression as a multiplication: \[ (\sin x + \cos x) \cdot \frac{\cos x}{\cos x+\sin x}. \] **Step 5.** Notice that \(\sin x+\cos x\) appears in both the numerator and the denominator. Cancel these common factors (assuming \(\sin x+\cos x \neq 0\)): \[ \cos x. \] Thus, we have shown that \[ \frac{\sin x+\cos x}{1+\tan x} = \cos x. \]

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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.

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