I'm so lost on this trigonometry problem :(?

Question

Evaluate: tan 13pi/12 (Use the fact that 13pi/12 =4pi/3- pi/4)

Answer 1

tan(a-b)=(tana-tanb)/(1+tan a*tan b)
so,
tan 13pi/12=tan(4pi/3-pi/4)
=(tan4pi/3-tanpi/4)/
(1+tan4pi/3*tanpi/4)
=(sqrt3-1)/(1+sqrt3)
=(sqrt3-1)^2/3-1
=(4-2sqrt3)/2
=2-sqrt3

Answer 2

You will need to use the sun and difference identities. If i were better at trig, I would do the problem, but I'm not good, so I'll give you an equation: tan(alpha+-beta) = (tan(alpha)+-tan(beta)) / (1-+tan(alpha)(tan(beta)))
where (alpha) and (beta) are the measures of 2 angles

***** note that the +- and -+ symbolize what sign should be used in different situations.
(It is actually written with the + above the - and the - above the +, respectively, but there isn't a key on the keyboard for that)
Example:
If the left hand side (LHS) of the equation used + (alpha+beta), then the sign in the numerator of the right hand side(RHS) of the equation needs to be the same, hence the +-, and the denominator needs to be opposite, hence the -+

so if its tan (30 degrees + 5 degrees) then the numerator needs to use the + sign and the denominator needs to use the - sign.
if: tan (30 - 5) then the numerator needs to use the - sign and the denominator needs to use the + sign.

I hope this makes sence :/

Answer 3

tan(13pi/12) = tan((4pi/3) - (pi/4))

tan((4pi/3) - (pi/4)) = (tan(4pi/3) - tan(pi/4))/(1 + (tan(4pi/3)tan(pi/4)))

tan((4pi/3) - (pi/4)) = (sqrt(3) - 1)/(1 + (sqrt(3) * 1))
sqrt((4pi/3) - (pi/4)) = (sqrt(3) - 1)/(1 + sqrt(3))
sqrt((4pi/3) - (pi/4)) = (sqrt(3) - 1)/(sqrt(3) + 1))

Multiply top and bottom by sqrt(3) - 1

sqrt((4pi/3) - (pi/4)) = ((sqrt(3) - 1)(sqrt(3) - 1))/(3 - 1)
sqrt((4pi/3) - (pi/4)) = (3 - sqrt(3) - sqrt(3) + 1)/2
sqrt((4pi/3) - (pi/4)) = (4 - 2sqrt(3))/2
sqrt((4pi/3) - (pi/4)) = 2 - sqrt(3)

ANS : tan(13pi/12) = 2 - sqrt(3)

If you were to do tan(13pi/12) and 2 - sqrt(3), the will match.

Answer 4

you get tan4pi/3 - tan(pi/4)?