what is cot(-13π/6)?

Explanation:

Step1: Use cotangent - angle property

Since $\cot(-\alpha)=-\cot\alpha$, then $\cot(-\frac{13\pi}{6})=-\cot(\frac{13\pi}{6})$.

Step2: Reduce the angle

$\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$. And the cotangent function has a period of $\pi$, so $\cot(\frac{13\pi}{6})=\cot(2\pi + \frac{\pi}{6})=\cot(\frac{\pi}{6})$.

Step3: Recall the value of cotangent

We know that $\cot\theta=\frac{\cos\theta}{\sin\theta}$, and for $\theta = \frac{\pi}{6}$, $\sin(\frac{\pi}{6})=\frac{1}{2}$, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, so $\cot(\frac{\pi}{6})=\frac{\cos(\frac{\pi}{6})}{\sin(\frac{\pi}{6})}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$.

Step4: Find the original value

Since $\cot(-\frac{13\pi}{6})=-\cot(\frac{13\pi}{6})=-\cot(\frac{\pi}{6})$, then $\cot(-\frac{13\pi}{6})=-\sqrt{3}$.

Answer:

$3$ (because the expression is $-\sqrt{?}$ and the value of $\cot(-\frac{13\pi}{6})$ is $-\sqrt{3}$, so the number inside the square - root is $3$)