Question

rewrite (\frac{cos\theta - sin\theta}{sin\theta+cos\theta}) over a common denominator. type your answer in terms of sine and/or cosine

(\frac{cos\theta}{sin\theta}=square) (simplify your answer)

Explanation:

Step1: Find common denominator

The common denominator of the two fractions is $(\sin\theta+\cos\theta)(\cos\theta - \sin\theta)$.

Step2: Rewrite fractions with common denominator

$\frac{\cos\theta}{\sin\theta+\cos\theta}=\frac{\cos\theta(\cos\theta - \sin\theta)}{(\sin\theta+\cos\theta)(\cos\theta - \sin\theta)}$ and $\frac{\sin\theta}{\cos\theta - \sin\theta}=\frac{\sin\theta(\sin\theta+\cos\theta)}{(\sin\theta+\cos\theta)(\cos\theta - \sin\theta)}$.

Step3: Subtract the fractions

$\frac{\cos\theta(\cos\theta - \sin\theta)-\sin\theta(\sin\theta+\cos\theta)}{(\sin\theta+\cos\theta)(\cos\theta - \sin\theta)}=\frac{\cos^{2}\theta-\cos\theta\sin\theta-\sin^{2}\theta-\sin\theta\cos\theta}{(\sin\theta+\cos\theta)(\cos\theta - \sin\theta)}$.

Step4: Simplify the numerator

$\frac{\cos^{2}\theta-\sin^{2}\theta - 2\sin\theta\cos\theta}{(\sin\theta+\cos\theta)(\cos\theta - \sin\theta)}$. Using the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$ where $a=\cos\theta$ and $b = \sin\theta$, the denominator is $\cos^{2}\theta-\sin^{2}\theta$.

Step5: Simplify the fraction

$\frac{\cos^{2}\theta-\sin^{2}\theta - 2\sin\theta\cos\theta}{\cos^{2}\theta-\sin^{2}\theta}=1-\frac{2\sin\theta\cos\theta}{\cos^{2}\theta-\sin^{2}\theta}$

Answer:

$1-\frac{2\sin\theta\cos\theta}{\cos^{2}\theta - \sin^{2}\theta}$