Before reading the proof, you must be familiar with the addition and subtraction rule for cosine and sine. As for the proof, first lets start by dividing sin(j - k) by cos(j - k):
Now lets divide both the numerator and the denominator by cos(j)cos(k):
We already know that tan(x) is [sin(x)/cos(x)], so:
And that is the subtraction formula for tan(x) As for the addition rule, we do the same thing we did above but with sin(j + k) and cos(j + k) instead:
Now lets divide both the numerator and the denominator by cos(j)cos(k):
We already know that tan(x) is [sin(x)/cos(x)], so:
And that is the addition formula for tan(x). Now let's replace k with -z.
\[{\color{deepskyblue}\tan({\color{orchid}j} + {\color{green}-z})} = \frac{{\color{deepskyblue}\tan({\color{orchid}j})} + {\color{deepskyblue}\tan({\color{green}-z})}}{1- {\color{deepskyblue}\tan({\color{orchid}j}) \tan({\color{green}-z})}}\]
Since tan(-x) = -tan(x):
\[{\color{deepskyblue}\tan({\color{orchid}j} - {\color{green}z})} = \frac{{\color{deepskyblue}\tan({\color{orchid}j})} - {\color{deepskyblue}\tan({\color{green}z})}}{1 + {\color{deepskyblue}\tan({\color{orchid}j}) \tan({\color{green}z})}}\]