Find S = tan 3pi + cot ((11pi)/12)
Use trig table, unit circle, and property of complement arcs: -->
tan (3pi) = 0
cot ((11pi)/12) = cot ((5pi)/12 + pi/2) = - tan ((5pi)/12)
Evaluate tan ((5pi)/12) by applying trig identity:
tan 2a = (2tan a)/(1 - tan^2 a)
In this case, call tan ((5pi)/12) = tan t
tan (2t) = tan ((10pi)/12) = tan ((5pi)/6) = -1/sqrt3 = (2tan t)/(1 - tan^2 t)
Cross multiply, and solve the quadratic equation for tan t
tan^2 t - 2sqrt3tan t - 1 = 0
D = d^2 = b^2 - 4ac = 12 + 4 = 16 --> d = +- 4
tan t = -b/(2a) +- d/(2a) = 2/2 +- 4/2 = 1 +- 2
tan t = tan ((5pi)/12) = 3 and tan t = -1
Since tan ((5pi)/12) is positive, take the positive value (tan t = 3)
Finally,
S = 0 + cot ((11pi)/12) = 0 - tan ((5pi)/12) = 0 - 3 = - 3
