I was in between proving a trigonometric identity but couldn't succeed. I went through the solution and saw this in between

\begin{align}\frac{\cos A \cos B}{\sin A \sin B}&= \frac{3}{1}\\\\ \frac{\cos A \cos B +\sin A \sin B}{\cos A \cos B - \sin A \sin B}&= \frac{3+1}{3-1}\end{align}

What happened there in the second step?


If we have $$ \frac xy = \frac31 $$ then this means, by definition of fractions, that $x = 3y$. This yields $$ \frac{x+y}{x-y} =\frac{3y+y}{3y-y} = \frac{3+1}{3-1} $$ In your case, $x = \cos A\cos B$ and $y = \sin A\sin B$.


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