# One more trig. problem

Given that tan 75 = 2 + root 3, find in the form m + n root 3, where m and n are integers, the value of tan 15.

I can't get this one. There's another one which is tan 105 but I got that straight away, its - 2 - root 3. Can anyone tell me how to work out this problem please?
mik1a
Given that tan 75 = 2 + root 3, find in the form m + n root 3, where m and n are integers, the value of tan 15.

I can't get this one. There's another one which is tan 105 but I got that straight away, its - 2 - root 3. Can anyone tell me how to work out this problem please?

A start would be that a right angled triangle with another angle of 75 has third angle of 15, so begin by drawing said triangle...
so tan 75 would be the opposite side over the adjacent side. so i convert 2+root3 into a fraction... (3+2root3)/(root3) therefore tan 15 must be (root3) / (3+2root3) ?

I know that's wrong ...
Tan 15 = tan(75 - 60)
tan(75-60) = (tan 75 - tan 60)/(1 + tan75tan60)

tan60 = tan(30+30) = tan 30 + tan 30)/(1 - tan^2 30)
You should know that tan 30 = sqrt(1/3)

So tan 60 = 2sqrt(1/3)/(1-1/3) = 3sqrt(1/3) = sqrt(3).

tan 15 = (2 + sqrt 3 - sqrt3)/(1+3+2sqrt(3))
tan 15 = 2/(4+2sqrt3) = 1/(2 + sqrt3)

Rationalise the denominator to get
tan 15 = (2-sqrt3)/(4 - 3) = 2-sqrt3
One quick trick that i like is this.
tan(90 - x) = 1/(tan x)
This greatly simplifies your problem since 15 = 90-75

tan 90-75 = 1/tan75 = 1/2+sqrt3 = 2-sqrt3.
JamesF
One quick trick that i like is this.
tan(90 - x) = 1/(tan x)
This greatly simplifies your problem since 15 = 90-75

tan 90-75 = 1/tan75 = 1/2+sqrt3 = 2-sqrt3.

Nice trick, how does it apply to the other trigonometric functions?
Im not sure what you mean, it doesnt work for sin or cos. The way to prove it is
tan(90-x) = sin(90-x)/cos(90-x) = cosx/sinx = 1/tanx
JamesF
Im not sure what you mean, it doesnt work for sin or cos. The way to prove it is
tan(90-x) = sin(90-x)/cos(90-x) = cosx/sinx = 1/tanx

Oh OK. Cheers!