Question:

Define \(z=x+iy\). Prove the following equation

\[ \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \tan z \]

Answer:

We should use the following relationships:

\[  \sinh x = -i\sin ix   \]

\[  \cosh x = \cos ix   \]

\[  \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}  \]

\[  \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \]

Thus,

\[  \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \frac{\sin 2x + i(-i\sin i2y)}{\cos 2x + \cos i2y}  \]

\[                                     = \frac{\sin 2x + \sin i2y}{\cos 2x + \cos i2y}  \]

\[                                     = \frac{2\sin\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}{2\cos\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}  \]

\[                                     = \frac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)}  \]

\[                                     = \frac{\sin(x+iy)}{\cos(x+iy)}  \]

\[                                     = \tan(x+iy) = \tan z   \]

Now, it’s proven!