A function [tex]f(z)=f(x+iy)=u(x,y)+i v(x,y)[/tex] is analytic if the C-R conditions are satisfied:
[tex]\begin{cases}u_x=v_y\\u_y=-v_x\end{cases}[/tex]
We have
[tex]f(z)=\tan z=\tan(x+iy)[/tex]
Recall the angle sum formula for tangent:
[tex]\tan(x+iy)=\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}[/tex]
Now recall that
[tex]\tan iy=\dfrac{\sin iy}{\cos iy}=\dfrac{-\frac{e^y-e^{-y}}{2i}}{\frac{e^y+e^{-y}}2}=\dfrac{i\sinh y}{\cosh y}=i\tanh y[/tex]
So we have
[tex]\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}=\dfrac{\tan x+i\tanh y}{1-i\tan x\tanh y}[/tex]
[tex]=\dfrac{(\tan x+i\tanh y)(1+i\tan x\tanh y)}{(1-i\tan x\tanh y)(1+i\tan x\tanh y)}[/tex]
[tex]=\dfrac{\tan x+i\tanh y+i\tan^2x-\tan x\tanh^2y}{1+\tan^2x\tanh^2y}[/tex]
[tex]=\dfrac{\tan x(1-\tanh^2y)}{1+\tan^2x\tanh^2y}+i\dfrac{\tanh y(1+\tan^2x)}{1+\tan^2x\tanh^2y}[/tex]
[tex]=\underbrace{\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}}_{u(x,y)}+i\underbrace{\dfrac{\tanh y\sec^2x}{1+\tan^2x\tanh^2y}}_{v(x,y)}[/tex]
We could stop here, but taking derivatives may be messy and would be easier to do if we can write this in terms of sines and cosines.
[tex]u(x,y)=\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}=\dfrac{\frac{\sin x}{\cos x\cosh^2x}}{1+\frac{\sin^2x\sinh^2y}{\cos^2x\cosh^2y}}=\dfrac{\sin x\cos x}{\cos^2x\cosh^2y+\sin^2x\sinh^2y}[/tex]
[tex]u(x,y)=\dfrac{\sin2x}{2\cos^2x\cosh^2y+2(1-\cos^2x)(\cosh^2y-1)}=\dfrac{\sin2x}{2\cosh^2y-1+2\cos^2x-1}[/tex]
[tex]u(x,y)=\dfrac{\sin2x}{\cos2x+\cosh2y}[/tex]
With similar usage of identities, we can find that
[tex]v(x,y)=\dfrac{\sinh2y}{\cos2x+\cosh2y}[/tex]
Now we check the C-R conditions.
[tex]u_x=\dfrac{2\cos2x(\cos2x+\cosh2y)-(-2\sin2x)\sin2x}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cos2x\cosh2y}{(\cos2x+\cosh2y)^2}[/tex]
[tex]v_y=\dfrac{2\cosh2y(\cos2x+\cosh2y)-(2\sinh2y)\sinh2y}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cosh2y\cos2x}{(\cos2x+\cosh2y)^2}[/tex]
[tex]\implies u_x=v_y[/tex]
Similarly, you can check that [tex]u_y=-v_x[/tex], hence the C-R conditions are satisfied, and so [tex]\tan z[/tex] is analytic.
