Compound angle formula Vs Double angle formula

 Prove that $ \displaystyle \frac{{\cos 3A+\sin 3A}}{{\cos A-\sin A}}=1+2\sin 2A$ .

$ \displaystyle \begin{array}{l}\frac{{\cos 3A+\sin 3A}}{{\cos A-\sin A}}=\frac{{\cos (2A+A)+\sin (2A+A)}}{{\cos A-\sin A}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos 2A\cos A-\sin 2A\sin A+\sin 2A\cos A+\cos 2A\sin A}}{{\cos A-\sin A}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos 2A(\cos A+\sin A)+\sin 2A(\cos A-\sin A)}}{{\cos A-\sin A}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{({{{\cos }}^{2}}A-{{{\sin }}^{2}}A)(\cos A+\sin A)}}{{\cos A-\sin A}}+\sin 2A\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(\cos A+\sin A)}^{2}}+\sin 2A\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\cos }^{2}}A+2\sin A\cos A+{{\sin }^{2}}A+\sin 2A\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+\sin 2A+\sin 2A\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+2\sin 2A\end{array}$