Example  TRIG.INV.1:    Suppose $f(x)=2tan(x−1)−3$ for $x\in(−\fracπ2+1,\fracπ2+1)$. Verify that $g(x)=arctan(\frac{x+3}2)+1$ is the inverse function for $f$ .

(i)$(f \circ g)(x) = f(g(x)) =f( arctan(\frac{x+3}2)+1) = 2tan((arctan(\frac{x+3}2)+1) -1)- 3 = 2 \frac{x+3}2 -3= x$.
(ii)$(g \circ f)(x) = g(f(x)) =g(2tan(x−1)−3) = arctan (\frac {(2tan(x−1)−3)+3}2) + 1  = (x-1)+1 = x $.

We check both with Mapping Diagrams and the cartesian graphs..
Which do you find more convincing visually?

Select the point $x$ in the mapping diagram to move the point by mouse or use the scroll up or down key to move by increments.