# Question #b8ee8

Apr 5, 2017

(- 3)

#### Explanation:

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