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I was working through a trigonometry problem, and was having some difficulty so I decided to look at the solution. Here are the steps:

$$\frac{\sin(2x+50^\circ)+\sin(150^\circ)}{\sin(2x+50^\circ)-\sin(150^\circ)}=\frac{\cos(50^\circ)-\cos(2x+50^\circ)}{\cos(50^\circ)+\cos(2x+50^\circ)}$$ $$\frac{\sin(2x+50^\circ)}{\sin150^\circ}=\frac{-\cos50^\circ}{\cos(2x+50^\circ)}$$

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I am not exactly how the solution got from the first step to the second one. I would just like some clarification on the intermediate step.

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$$\frac{a+b}{a-b}=\frac{c-d}{c+d}$$

$$ac+ad+bc+bd=ac-ad-bc+bd$$

$$2ad=-2bc$$

$$\frac{a}{b}=-\frac{c}{d}$$

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This is similar to Siong's answer, but written differently using ratios: \begin{align} \frac{a+b}{a-b} - 1 &= \frac{c-d}{c+d} -1 \\[2ex] \require{cancel} \frac{2b}{a-b} &= -\frac{2d}{c+d} \\[2ex] \frac{a-b}{b} &= -\frac{c+d}{d} \\[2ex] \frac{a}{b} - 1 &= -\frac{c}{d} -1 \\ \end{align}

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