Trigonometric functions and complex variables

Question:
Define $z=x+iy$. Prove the following equation
\[ \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \tan z \]

Answer:
There are following relationships:
\begin{eqnarray*}
\sinh x &=& -i\sin ix   \\
\cosh x &=& \cos ix   \\
\sin A + \sin B &=& 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}  \\
\cos A + \cos B &=& 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} &=& \frac{\sin 2x + i(-i\sin i2y)}{\cos 2x + \cos i2y}\\
                                     &=& \frac{\sin 2x + \sin i2y}{\cos 2x + \cos i2y}  \\
                                     &=& \frac{2\sin\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}{2\cos\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}  \\
&=& \frac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)}  \\
&=& \frac{\sin(x+iy)}{\cos(x+iy)}  \\
&=& \tan(x+iy) = \tan z
\end{eqnarray*}

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