The exact value of $\cot\frac{7\pi}{6}$? I am working on a trigonometry question at the moment and am very stuck. I have looked through all the tips to solving it and I cant seem to come up with the right answer. The problem is 

What is exact value of
  $$\cot \left(\frac{7\pi}{6}\right)? $$

 A: We have $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{\cos \left(\frac{7\pi}{6}\right)}{\sin \left(\frac{7\pi}{6}\right)} \equiv \frac{-\cos \left(\frac{\pi}{6}\right)}{-\sin \left(\frac{\pi}{6}\right)} = \sqrt{3}$$
You can easily see this using a "CAST" diagram to reduce $\cos \left(\frac{7\pi}{6}\right)$and $\sin \left(\frac{7\pi}{6}\right)$ to standard results.

As per @Scientifica's comment, an easier method would be to simply note that $\tan$ is $\pi$-periodic so that shifting its argument by $\pi$ will yield no change to it's value, or: $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{1}{\tan \left(\frac{7\pi}{6} - \pi\right)} = \frac{1}{\tan \left(\frac{\pi}{6}\right)} = \sqrt{3}$$
A: $$\cot \left(\frac{7\pi}{6}\right) = \cot \left(\frac{\pi}{6}\right) = \sqrt 3 $$
because $$ \tan ( \theta + \pi) = \tan ( \theta ) $$
A: Hint: $\cot x=\dfrac{\cos x}{\sin x}$ and
$$
\cos \left(\dfrac{7  \pi}{6}\right)=\cos \left(\pi+\dfrac{\pi}{6}\right)
$$
$$
\sin \left(\dfrac{7  \pi}{6}\right)=\sin \left(\pi+\dfrac{\pi}{6}\right)
$$
now you can use sum formulas or reduction to the first quadrant.
