cot ((pi)/12) = 1/(tan ((pi)/12)cot(π12)=1tan(π12). First find tan ((pi)/12)tan(π12)
Call tan ((pi)/12) = ttan(π12)=t
tan (2t) = tan ((pi)/6) = 1/sqrt3tan(2t)=tan(π6)=1√3
Apply the trig identity: tan 2a = (2tan a)/(1 - tan^2 a)tan2a=2tana1−tan2a
We get:
1/sqrt3 = (2t)/(1 - t^2)1√3=2t1−t2
1 - t^2 = 2sqrt3t1−t2=2√3t. Solve the quadratic equation in t.
t^2 + 2sqrt3t - 1 = 0t2+2√3t−1=0
D = d^2 = b^2 - 4ac = 12 + 4 = 16 D=d2=b2−4ac=12+4=16--> d = +- 4d=±4
There are 2 real roots:
tan ((pi)/12) = t = - 2sqrt3/2 +- 4/2 = -sqrt3 +- 2tan(π12)=t=−2√32±42=−√3±2
Since the arc (pi/12) is in Quadrant I, its tan is positive, then
tan ((pi)/12) = (-sqrt3 + 2).tan(π12)=(−√3+2).
Check by calculator:
tan ((pi)/12) = tan 15 = 0.27.tan(π12)=tan15=0.27.
(2 - sqrt3) = 0.27(2−√3)=0.27. OK
