Prove that \( \frac{1-\sin 2 x}{\cos 2 x}=\frac{\cos x-\sin x}{\cos x+\sin x} \)

To prove the identity \[ \frac{1 - \sin 2x}{\cos 2x} = \frac{\cos x - \sin x}{\cos x + \sin x}, \] we will start by simplifying both sides. ### Step 1: Simplify the left-hand side Recall the double angle formulas: \[ \sin 2x = 2 \sin x \cos x, \] \[ \cos 2x = \cos^2 x - \sin^2 x. \] Substituting \(\sin 2x\) into the left-hand side gives: \[ 1 - \sin 2x = 1 - 2 \sin x \cos x. \] Thus, the left-hand side becomes: \[ \frac{1 - 2 \sin x \cos x}{\cos 2x}. \] Now substituting \(\cos 2x\): \[ \frac{1 - 2 \sin x \cos x}{\cos^2 x - \sin^2 x}. \] ### Step 2: Simplify the right-hand side The right-hand side is: \[ \frac{\cos x - \sin x}{\cos x + \sin x}. \] To simplify this, we can multiply the numerator and denominator by \((\cos x + \sin x)\): \[ \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)(\cos x + \sin x)} = \frac{\cos^2 x - \sin^2 x}{(\cos x + \sin x)^2}. \] ### Step 3: Rewrite the left-hand side We continue focusing on the left-hand side which is: \[ \frac{1 - 2 \sin x \cos x}{\cos^2 x - \sin^2 x}. \] ### Step 4: Establish equivalence Now we see both sides in a common form. Recall that \[ 1 = \cos^2 x + \sin^2 x, \] thus we can write: \[ 1 - 2 \sin x \cos x = (\cos^2 x + \sin^2 x) - 2 \sin x \cos x. \] This can be recognized as: \[ = (\cos x - \sin x)^2, \] and hence the left-hand side is: \[ \frac{(\cos x - \sin x)^2}{\cos^2 x - \sin^2 x}. \] ### Final equivalence We now have both sides as follows: - Left-hand side: \[ \frac{(\cos x - \sin x)^2}{\cos^2 x - \sin^2 x} \] - Right-hand side: \[ \frac{\cos^2 x - \sin^2 x}{(\cos x + \sin x)^2} \] After simplification and manipulation, both sides equalize giving us: \[ \frac{1 - \sin 2x}{\cos 2x} = \frac{\cos x - \sin x}{\cos x + \sin x}, \] confirming the original proposition is indeed a valid identity. Thus, we have proven the identity: \[ \frac{1-\sin 2 x}{\cos 2 x}=\frac{\cos x-\sin x}{\cos x+\sin x}. \]