Question 277603
Try:
To make this more natural, remember that the primitive tangent function has period of pi. So we have Tan(-5pi/12) congruent to Tan(7pi/12) mod pi.

Now: What functions do we easily know the value to for trigonometrics? pi/3 is certainly one. So, pi/3=4pi/12---> We need to know 3pi/12=pi/4 ! This is also an easy find.

So, we take Tan(7pi/12)=Tan(3pi/12   +  4pi/12) Now we use your formulae:

Tan(3pi/12   +  4pi/12)=[Tan(pi/4)-Tan(pi/3)]/(1+Tan(pi/4)Tan(pi/3))

Now it remains to remember what Tan(pi/4) and Tan(pi/3) are. We use the definition of Tangent.
Tan(x)=Sin(x)/Cos(x).
So Tan(pi/4)=1.
Tan(pi/3)=(sqrt(3)/2)/(1/2)=sqrt(3)

Finally, substitute into our expression:

(1-sqrt(3))/(1+sqrt(3))

Now, simplify:

{{{(1-sqrt(3))^2/(1-3)}}}={{{(1-2*sqrt(3)+3)*(1/(-2))}}}={{{-2-sqrt(3)}}}

This could be done quicker, but I hope you can see the full details of what is going on through this procedure.