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\( \frac { \cos x + \sin x } { \cos x - \sin x } - \frac { \cos x - \sin x } { \cos x + \sin x } = 2 \tan 2 x \)

Ask by Goodwin Mitchell. in South Africa
Mar 09,2025

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The equation simplifies to \(2 \tan 2x\).

Solución

Given the equation: \[ \frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} = 2 \tan 2x \] To simplify the left side of the equation, we can use the trigonometric identity for the tangent of a sum of angles: \[ \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \] Let's simplify the left side of the equation using this identity. Simplify the expression by following steps: - step0: Solution: \(\frac{\cos\left(x\right)+\sin\left(x\right)}{\cos\left(x\right)-\sin\left(x\right)}-\frac{\cos\left(x\right)-\sin\left(x\right)}{\cos\left(x\right)+\sin\left(x\right)}\) - step1: Reduce fractions to a common denominator: \(\frac{\left(\cos\left(x\right)+\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}-\frac{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)-\sin\left(x\right)\right)}{\left(\cos\left(x\right)+\sin\left(x\right)\right)\left(\cos\left(x\right)-\sin\left(x\right)\right)}\) - step2: Rewrite the expression: \(\frac{\left(\cos\left(x\right)+\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}-\frac{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)-\sin\left(x\right)\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}\) - step3: Transform the expression: \(\frac{\left(\cos\left(x\right)+\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)-\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)-\sin\left(x\right)\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}\) - step4: Multiply the terms: \(\frac{\cos^{2}\left(x\right)+2\cos\left(x\right)\sin\left(x\right)+\sin^{2}\left(x\right)-\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)-\sin\left(x\right)\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}\) - step5: Multiply the terms: \(\frac{\cos^{2}\left(x\right)+2\cos\left(x\right)\sin\left(x\right)+\sin^{2}\left(x\right)-\left(\cos^{2}\left(x\right)-2\cos\left(x\right)\sin\left(x\right)+\sin^{2}\left(x\right)\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}\) - step6: Calculate: \(\frac{4\cos\left(x\right)\sin\left(x\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}\) - step7: Transform the expression: \(\frac{4\cos\left(x\right)\sin\left(x\right)}{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)}\) - step8: Transform the expression: \(\frac{2\sin\left(2x\right)}{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)}\) - step9: Transform the expression: \(\frac{2\sin\left(2x\right)}{2\cos^{2}\left(x\right)-1}\) - step10: Transform the expression: \(\frac{4\cos\left(x\right)\sin\left(x\right)}{2\cos^{2}\left(x\right)-1}\) - step11: Transform the expression: \(\frac{4\cos\left(x\right)\sin\left(x\right)}{1-2\sin^{2}\left(x\right)}\) - step12: Transform the expression: \(\frac{4\cos\left(x\right)\sin\left(x\right)}{-1+2\cos^{2}\left(x\right)}\) - step13: Transform the expression: \(\frac{4\cos\left(x\right)\sin\left(x\right)}{\cos\left(2x\right)}\) - step14: Rewrite the expression: \(\frac{2\sin\left(2x\right)}{\cos\left(2x\right)}\) - step15: Rewrite the expression: \(2\tan\left(2x\right)\) The simplified form of the left side of the equation is \(2 \tan 2x\). Therefore, the given equation is true, and we can conclude that: \[ \frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} = 2 \tan 2x \]

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Bonus Knowledge

Let’s simplify the left-hand side of the equation: \[ LHS = \frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} \] To combine these two fractions, we need a common denominator, which is \((\cos x - \sin x)(\cos x + \sin x)\). Therefore, \[ LHS = \frac{(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)} \] Now, let's expand the numerators: 1. \((\cos x + \sin x)^2 = \cos^2 x + 2\cos x \sin x + \sin^2 x = 1 + 2\cos x \sin x\) 2. \((\cos x - \sin x)^2 = \cos^2 x - 2\cos x \sin x + \sin^2 x = 1 - 2\cos x \sin x\) Now substitute back into the expression: \[ LHS = \frac{(1 + 2\cos x \sin x) - (1 - 2\cos x \sin x)}{(\cos x - \sin x)(\cos x + \sin x)} \] Simplifying the numerator: \[ LHS = \frac{4\cos x \sin x}{(\cos x - \sin x)(\cos x + \sin x)} \] Recall that \(\cos x + \sin x = \sqrt{2} \sin(x + \frac{\pi}{4})\) and \(\cos x - \sin x = \sqrt{2} \cos(x + \frac{\pi}{4})\). Therefore, \[ LHS = \frac{4\cos x \sin x}{\cos^2 x - \sin^2 x} = \frac{2\sin 2x}{\cos 2x} \] Now, let's rewrite our expression: \[ LHS = 2 \tan 2x \] This confirms that: \[ \frac { \cos x + \sin x } { \cos x - \sin x } - \frac { \cos x - \sin x } { \cos x + \sin x } = 2 \tan 2x \] Thus, the equation is verified.

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