Let f(x)=(x-2)/2, then in [0, π]

$x \in [0, \pi] \Rightarrow \frac{x-2}{2} \in\left[-1, \frac{\pi}{2}-1\right]$

Now $\frac{1}{f(x)}=\frac{2}{x-2}$ which is discontinuations at x = 2.  tan (f(x)) is continuous for $x \in$ $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) . f^1(x)=2(x+1)$ 

which is clearly continuous but $\tan \left(f^{-1}(x)\right)$ is not continuous.

Hence (2) is the correct answer.