# Question #d6fb5

Dec 4, 2015

See explanation.

#### Explanation:

$\left[1\right] \text{ } = \csc \left(\frac{\pi}{12}\right)$

Reciprocal Identity: $\csc \theta = \frac{1}{\sin} \theta$

$\left[2\right] \text{ } = \frac{1}{\sin} \left(\frac{\pi}{12}\right)$

Represent $\frac{\pi}{12}$ as a difference of two special angles.

$\left[3\right] \text{ } = \frac{1}{\sin} \left(\frac{3 \pi}{12} - \frac{2 \pi}{12}\right)$

$\left[4\right] \text{ } = \frac{1}{\sin} \left(\frac{\pi}{4} - \frac{\pi}{6}\right)$

Difference Identity: $\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

$\left[5\right] \text{ } = \frac{1}{\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) - \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right)}$

You can solve these since $\frac{\pi}{4}$ and $\frac{\pi}{6}$ are special angles.

$\left[6\right] \text{ } = \frac{1}{\left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)}$

$\left[7\right] \text{ } = \frac{1}{\left(\frac{\sqrt{6}}{4}\right) - \left(\frac{\sqrt{2}}{4}\right)}$

$\left[8\right] \text{ } = \frac{1}{\frac{\sqrt{6} - \sqrt{2}}{4}} \cdot \frac{4}{4}$

$\left[9\right] \text{ } = \frac{4}{\sqrt{6} - \sqrt{2}}$

Rationalize the denominator.

$\left[10\right] \text{ } = \frac{4}{\sqrt{6} - \sqrt{2}} \cdot \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}}$

$\left[11\right] \text{ } = \frac{4 \left(\sqrt{6} + \sqrt{2}\right)}{6 - 2}$

$\left[12\right] \text{ } = \frac{\cancel{4} \left(\sqrt{6} + \sqrt{2}\right)}{\cancel{4}}$

$\left[13\right] \text{ } = \textcolor{b l u e}{\sqrt{6} + \sqrt{2}}$