Is $\tan(x)$ differentiable for $x\in ( -\pi/2 , \pi/2 )$ This is an assignment question and in class we taught the definition that:

A function $f(x)$ is differentiable if we can find $f(x+h) - f(x) = Kh +h E(x,h)$, where $K=f'(x)$ and $E(x,h) \rightarrow 0$   as $h \rightarrow 0$
E.g. $f(x) = x^3 \Rightarrow f(x+h) - f(x) = (x+h)^3 - x^3 = 3x^2h+2xh^2 +h^3$
  and so $3x^2h+2xh^2+h^3 =Kh +h E(x,h) $. Thus we can conclude that $K = 3x^2$ and $E(x,h) = 2xh + h^2 $, and thus $f$ is differentiable since $E(x,h) = 2xh + h^2 \rightarrow 0$   as $h \rightarrow 0$

But for proving that $tan(x)$ is differentiable I started and got stuck:
$\tan(x+h) - \tan(x) = \dfrac{\tan(h)(1+\tan^2(x))}{1-\tan(x)\tan(h)}$, I really want to find $K = 1+ \tan^2(x)$ but I can't find a way to get $(1+\tan^2(x))h$ without a fraction term, and $E(x,h) = 0 , h \rightarrow 0$.
Any hints or help would be great, thanks.

You may exploit the identity:
$$\tan(x+h)-\tan(x)=\frac{\sin h}{\cos(x+h)\cos(x)}$$
from which it follows that:
$$|\tan(x+h)-\tan(x)|\leq\frac{|h|}{\min(\cos x,\cos(x+h))^2}.\tag{1}$$
Now the denominator of the RHS of $(1)$ may be very close to zero, but is never zero on $I=(-\pi/2,\pi/2)$, hence your function is differentiable over $I$.
