Recall that 15˚ can be written as 60˚ - 45˚.
So, tan15˚ = tan(60˚ - 45˚). We have to expand this using the difference formula for tangent. The sum/difference formulas are shown in the following table.
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tan(60˚ - 45˚) = (tan60˚- tan45˚)/(1 + tan60˚tan45˚)
tan(60˚ - 45˚) = (sqrt(3) - 1)/(1 + sqrt(3) xx 1)
tan(15˚) = (sqrt(3) - 1)/(1 + sqrt(3)
It is preferable that the denominator is rationalized.
tan15˚ = (sqrt(3) - 1)/(1 + sqrt(3)) xx (1 - sqrt(3))/(1 - sqrt(3))
tan15˚ = (sqrt(3) + sqrt(3) - 1 - sqrt(9))/(1 - sqrt(9))
tan15˚ = (2sqrt(3) - 4)/(-2)
tan15˚ = (2(sqrt(3) - 2))/-2
tan15˚ = -(sqrt(3) - 2)
tan15˚ = 2 - sqrt(3)
Hopefully this helps!
