Question

How do you use the angle sum or difference identity to find the exact value of `(tan(pi/4)-tan(pi/6))/(1+tan(pi/4)tan(pi/6))`?

Answer 1

`2-sqrt3`

Explanation:

Since

`tan(A-B) = {tanA-tanB}/{1+tanAtanB}`

the expression is obviously equal to

`tan(pi/4-pi/6) = tan(pi/12)`

To find the "exact form" of `tan(pi/12)` we use a couple of double angle formulas (which, of course are special cases of sum angle formulas) :

`tan(pi/12) = sin(pi/12)/cos(pi/12) = {2sin(pi/12)cos(pi/12)}/{2cos^2(pi/12)}=sin(pi/6)/{cos(pi/6)+1}={1/2}/{sqrt3/2+1}=1/{2+sqrt3}=2-sqrt3`