# Question #0280a

Oct 1, 2016

$\sin \left(\theta\right) = \frac{6 \sqrt{52}}{52}$, $\cos \left(\theta\right) = \frac{- 4 \sqrt{52}}{52}$, $\sec \left(\theta\right) = - \frac{\sqrt{52}}{4}$, $\csc = \frac{\sqrt{52}}{6}$ and $\cot \left(\theta\right) = - \frac{4}{6}$

#### Explanation:

The cotangent is the reciprocal of the tangent:

$\cot \left(\theta\right) = - \frac{4}{6}$

Use $1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$ to obtain the secant and the cosine; we know that both the secant and the cosine must be negative because we are given $\sin \left(\theta\right) > 0$

$1 + {\left(\frac{- 6}{4}\right)}^{2} = {\sec}^{2} \left(\theta\right)$

${\sec}^{2} \left(\theta\right) = \frac{16}{16} + \frac{36}{16}$

${\sec}^{2} \left(\theta\right) = \frac{52}{16}$

$\sec \left(\theta\right) = - \frac{\sqrt{52}}{4}$

$\cos \left(\theta\right) = \frac{- 4 \sqrt{52}}{52}$

$\sin \left(\theta\right) = \left(- \frac{6}{4}\right) \cos \left(\theta\right)$

$\sin \left(\theta\right) = \frac{6 \sqrt{52}}{52}$

$\csc = \frac{\sqrt{52}}{6}$