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OpenStudy (anonymous):
Prove that: \[\frac{\cos3A-cosA}{\sin3A-sinA}+\frac{\cos2A-\cos4A}{\sin4A-\sin2A}=\frac{sinA}{\cos2Acos3A}\]
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OpenStudy (experimentx):
first of all simplify using sin C - Sin D formula
OpenStudy (anonymous):
what math is this?
OpenStudy (experimentx):
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