It's a good idea in this case to rewrite $\tan A=\frac{\sin A}{\cos A}$, then simplify the expression on the right, multiplying top and bottom by $\cos A$ to clear the "little fractions" from the "big fraction".
That will give you $\frac{\cos A-\sin A}{\cos A+\sin A}$ on the right hand side. Then, rewriting the left hand side denominator as $(\cos A+\sin A)(\cos A-\sin A)$--as you already have--it remains only to somehow rewrite $1-2\sin A\cos A$ as $(\cos A-\sin A)^2$, which can be done by using the Pythagorean identity.