To solve the cotangent of these angles, we can use the following identity for cotangent:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

(a) $\cot\left(\frac{5}{6} \pi\right)$

1. Find the reference angle:
   $\frac{5}{6} \pi = 150^\circ$.
2. Tangent of 150°:
   Since $\tan(150^\circ) = -\tan(30^\circ) = -\frac{1}{\sqrt{3}}$, the cotangent is:
   $\cot\left(\frac{5}{6} \pi\right) = \frac{1}{-\frac{1}{\sqrt{3}}} = -\sqrt{3}$ So, $\cot\left(\frac{5}{6} \pi\right) = -\sqrt{3}$.

(b) $\cot\left(-\frac{2}{3} \pi\right)$

1. Find the reference angle:
   $-\frac{2}{3} \pi = -120^\circ$. In the third quadrant, where $-120^\circ$ lies, tangent is positive.
2. Tangent of 120°:
   $\tan(120^\circ) = -\tan(60^\circ) = -\sqrt{3}$. Therefore, the cotangent is:
   $\cot\left(-\frac{2}{3} \pi\right) = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}$ So, $\cot\left(-\frac{2}{3} \pi\right) = -\frac{1}{\sqrt{3}}$.

Final answers:

• (a) $\cot\left(\frac{5}{6} \pi\right) = -\sqrt{3}$
• (b) $\cot\left(-\frac{2}{3} \pi\right) = -\frac{1}{\sqrt{3}}$

Would you like further details on any step, or have any questions?

5 Related Questions:

1. What is the relationship between cotangent and sine/cosine?
2. How can we find the reference angle of a negative angle?
3. What is the value of $\cot(\frac{7}{6}\pi)$?
4. How can we determine the sign of trigonometric functions in different quadrants?
5. What is the exact value of $\tan\left(\frac{11}{6}\pi\right)$?

Tip:

Always remember to determine in which quadrant the angle lies, as it affects the sign of trigonometric functions.