<s>[INST] Trigonometric Simplification Question 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.
 [/INST] 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}
</s>