Let's consider the following integral $$\int \limits_{B(0, \pi / 4)} \frac{1}{z \tan(z)} \, \mbox{d}z.$$ Of course the only ,,hot'' point here would be $z_0 = 0$. The other ones are not inside the ball.
That implies $$\int \limits_{B(0, \pi / 4)} \frac{1}{z \tan(z)} \, \mbox{d}z = \pi \, i \,\text{res}_{z_0}.$$
How ever I can't figure out how to evaluate the limit. After differentiating the function twice I use l'Hospital rule once and I got $"1/0"$. I think that the result should be equal to $-2/3$. I would appreciate any hints or tips.
Since $\frac1{z\tan(z)}$ is an even function, it's residue at $0$ (which is the coefficient of $\frac1z$ in its Laurent series centered at $0$) is $0$. Therefore, your integral is equal to $0$.