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\frac{Sin\ 2a\ +\ Sin\ 2b}{Cos\ a\ -\ Cos\ 2b}\\\\=\frac{2\ Sin(\frac{2a+2b}{2})\ Cos(\frac{2a-2b}{2})}{2\ Sin(\frac{a+2b}{2})\ Sin(\frac{a-2b}{2})}\\\\=\frac{Sin(a+b)\ Cos(a-b)}{Sin(\frac{a}{2}+b)\ Sin(\frac{a}{2}-b)}\\\\=\frac{(Sina\ Cosb+Cosa\ sinb)(cosa\ cosb+sina\ sinb)}{}

There is some problem with the question given.  it is not right...
for example  a = 45 deg  and b = 45 deg.
LHS  is not equal to RHS

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