Find the principal value of `cot^(-1)(-sqrt3)`

Answer

To find the principal value of \( \cot^{-1}(-\sqrt{3}) \), we can follow these steps:

### Step 1: Identify the value of \( \sqrt{3} \) in terms of cotangent.
We know that:
\[ \cot\left(\frac{\pi}{6}\right) = \sqrt{3} \]
Thus, we can express \( -\sqrt{3} \) as:
\[ \cot\left(\frac{\pi}{6}\right) = \sqrt{3} \implies \cot\left(-\frac{\pi}{6}\right) = -\sqrt{3} \]

### Step 2: Use the cotangent identity.
Using the property of cotangent, we have:
\[ \cot(\pi - x) = -\cot(x) \]
So, we can write:
\[ \cot^{-1}(-\sqrt{3}) = \cot^{-1}(\cot(\pi - \frac{\pi}{6})) \]

### Step 3: Simplify the expression.
Now, we can simplify:
\[ \pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6} \]
Thus, we have:
\[ \cot^{-1}(-\sqrt{3}) = \cot^{-1}(\cot(\frac{5\pi}{6})) \]

### Step 4: Apply the inverse cotangent function.
Since \( \cot^{-1}(\cot(x)) = x \) for \( x \) in the range of \( (0, \pi) \), we can conclude:
\[ \cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6} \]

### Step 5: Verify the range.
The value \( \frac{5\pi}{6} \) is indeed within the principal range of \( (0, \pi) \).

### Final Answer:
Thus, the principal value of \( \cot^{-1}(-\sqrt{3}) \) is:
\[ \frac{5\pi}{6} \]

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