Cauchy Riemann of $\tan(z)$ So, I have to see if the Cauchy-Riemann condition holds for $\tan(z)$. To do this I have to get it into the form $u+iv$
This is what I have so far,
$$\tan(z) = \tan(x+iy) = \frac{\tan(x)+\tan(iy)}{1-\tan(x)\tan(iy)}.$$
Then I tried setting $\tan(x) = \frac{\sin(2x)}{1+\cos(2x)}$ and similarly for $\tan(iy)$ but I get stuck. So a hint on how to continue from $\frac{\tan(x)+\tan(iy)}{1-\tan(x)\tan(iy)}$ would be greatly appreciated.

Note that $\tan(iy) = \frac{\sin(iy)}{\cos(iy)}$. We know that
$$\sin(iy) = \frac{1}{2i}(e^{-y}-e^y) = i\sinh(y).$$
Likewise,
$$\cos(iy) = \frac{1}{2}(e^{-y}+e^y) = \cosh(y).$$
So we have that
$$\tan(z) = \frac{\tan(x)+i\tanh(y)}{1-i\tan(x)\tanh(y)}.$$
Can you see where to go from here? (Hint: try multiplying by a clever form of $1$...)
