Find the principal value of `sin^(-1)(-1/sqrt2)-2tan^(-1)(-sqrt3)+cos^(-1)(-1/2)`.

Answer

To find the principal value of the expression \( \sin^{-1}(-\frac{1}{\sqrt{2}}) - 2\tan^{-1}(-\sqrt{3}) + \cos^{-1}(-\frac{1}{2}) \), we can follow these steps:

### Step 1: Evaluate \( \sin^{-1}(-\frac{1}{\sqrt{2}}) \)
Using the property of inverse sine, we have:
\[ \sin^{-1}(-x) = -\sin^{-1}(x) \]
Thus,
\[ \sin^{-1}(-\frac{1}{\sqrt{2}}) = -\sin^{-1}(\frac{1}{\sqrt{2}}) \]
Since \( \sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4} \), we get:
\[ \sin^{-1}(-\frac{1}{\sqrt{2}}) = -\frac{\pi}{4} \]

### Step 2: Evaluate \( -2\tan^{-1}(-\sqrt{3}) \)
Using the property of inverse tangent, we have:
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
Thus,
\[ -2\tan^{-1}(-\sqrt{3}) = -2(-\tan^{-1}(\sqrt{3})) = 2\tan^{-1}(\sqrt{3}) \]
Since \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \), we get:
\[ -2\tan^{-1}(-\sqrt{3}) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 3: Evaluate \( \cos^{-1}(-\frac{1}{2}) \)
Using the property of inverse cosine, we know:
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
Thus,
\[ \cos^{-1}(-\frac{1}{2}) = \pi - \cos^{-1}(\frac{1}{2}) \]
Since \( \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \), we get:
\[ \cos^{-1}(-\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 4: Combine all parts
Now we combine all the evaluated parts:
\[ \sin^{-1}(-\frac{1}{\sqrt{2}}) - 2\tan^{-1}(-\sqrt{3}) + \cos^{-1}(-\frac{1}{2}) = -\frac{\pi}{4} + \frac{2\pi}{3} + \frac{2\pi}{3} \]
This simplifies to:
\[ -\frac{\pi}{4} + \frac{4\pi}{3} \]

### Step 5: Find a common denominator and simplify
The common denominator for \( 4 \) and \( 3 \) is \( 12 \):
\[ -\frac{\pi}{4} = -\frac{3\pi}{12} \]
\[ \frac{4\pi}{3} = \frac{16\pi}{12} \]
Thus, we have:
\[ -\frac{3\pi}{12} + \frac{16\pi}{12} = \frac{13\pi}{12} \]

### Final Answer
The principal value of the expression is:
\[ \frac{13\pi}{12} \]