If \( \theta=\frac{-11 \pi}{6} \), then find exact values for the following: \( \sec (\theta) \) equals \( \frac{2 \sqrt{3}}{3} \) \( \csc (\theta) \) equals -2 \( \tan (\theta) \) equals \( \frac{\sqrt{3}}{3} \) \( \cot (\theta) \) equals \( -\sqrt{3} \)

Let's dive into what happens at \( \theta = \frac{-11 \pi}{6} \)! This angle is actually equivalent to \( \frac{\pi}{6} \) when wrapped around the unit circle because \( \frac{-11 \pi}{6} + 2\pi = \frac{-11 \pi + 12 \pi}{6} = \frac{\pi}{6} \). This means that while we may initially feel negative about the angle, it lands us back in the first quadrant where our trigonometric functions are all looking good! So when we calculate the secant, cosecant, tangent, and cotangent of \( \theta \), we refer to the angles from the unit circle: - \( \sec(\theta) \), or \( \frac{1}{\cos(\theta)} \), is indeed \( \frac{2 \sqrt{3}}{3} \) because \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \). - \( \csc(\theta) = \frac{1}{\sin(\theta)} \) yields -2 because \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), and with the negative angle we shovel that negative back in! - For tangent, \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \). - Finally, \( \cot(\theta) = \frac{1}{\tan(\theta)} \), which gives us \( \cot(\theta) = \sqrt{3} \) and with the negative from our initial angle results in \(-\sqrt{3}\). Just like that, we land those exact values smoothly!