Find S = tan 3pi + cot ((11pi)/12)S=tan3π+cot(11π12)
Use trig table, unit circle, and property of complement arcs: -->
tan (3pi) = 0tan(3π)=0
cot ((11pi)/12) = cot ((5pi)/12 + pi/2) = - tan ((5pi)/12)cot(11π12)=cot(5π12+π2)=−tan(5π12)
Evaluate tan ((5pi)/12)tan(5π12) by applying trig identity:
tan 2a = (2tan a)/(1 - tan^2 a)tan2a=2tana1−tan2a
In this case, call tan ((5pi)/12) = tan ttan(5π12)=tant
tan (2t) = tan ((10pi)/12) = tan ((5pi)/6) = -1/sqrt3 = (2tan t)/(1 - tan^2 t)tan(2t)=tan(10π12)=tan(5π6)=−1√3=2tant1−tan2t
Cross multiply, and solve the quadratic equation for tan t
tan^2 t - 2sqrt3tan t - 1 = 0tan2t−2√3tant−1=0
D = d^2 = b^2 - 4ac = 12 + 4 = 16D=d2=b2−4ac=12+4=16 --> d = +- 4d=±4
tan t = -b/(2a) +- d/(2a) = 2/2 +- 4/2 = 1 +- 2tant=−b2a±d2a=22±42=1±2
tan t = tan ((5pi)/12) = 3 and tan t = -1tant=tan(5π12)=3andtant=−1
Since tan ((5pi)/12)tan(5π12) is positive, take the positive value (tan t = 3)
Finally,
S = 0 + cot ((11pi)/12) = 0 - tan ((5pi)/12) = 0 - 3 = - 3S=0+cot(11π12)=0−tan(5π12)=0−3=−3
