Question

How do you use the angle sum or difference identity to find the exact value of `tan(pi/6-pi/3)`?

Answer 1

`= -1/sqrt3`

Explanation:

FACTS TO KNOW:
`tan(pi/6) = 1/sqrt3`
`tan(pi/3) = sqrt3`
(The two above can be verified by drawing a 30-60-90 triangle)
`tan(A-B) = (tanA-tanB)/(1+tanAtanB)`

Therefore:
`tan(pi/6-pi/3) = (tan(pi/6)-tan(pi/3))/(1+tan(pi/6)tan(pi/3))`
`= (1/sqrt3-sqrt3)/(1+(1/sqrt3)sqrt3)`
`= (1/sqrt3-sqrt3)/(1+1)`
`= (1/sqrt3-sqrt3)/(2)`, then multiply top and bottom by `sqrt3`
`= (1-3)/(2sqrt3)`
`= -2/(2sqrt3)`
`= -1/sqrt3`

And thus this is how to use the difference/sum of angles formula

SOMETHING ELSE:
You can use the identity `tan(-x) = -tan(x)` to make this much simpler
`tan(pi/6-pi/3) = tan(-pi/3)`
`=-tan(pi/3)`
`=-1/sqrt3`