The value of `sin ^(-1) (-(1)/(sqrt2)) + cos ^(-1) (-(1)/(2)) - tan ^(-1) (-sqrt3) + cot ^(-1) (-(1)/(sqrt3)) ` is

Options

A. `(5pi)/(12)`
B. `(17pi)/(12)`
C. 0
D. 1

Answer

To solve the expression \( \sin^{-1} \left(-\frac{1}{\sqrt{2}}\right) + \cos^{-1} \left(-\frac{1}{2}\right) - \tan^{-1} \left(-\sqrt{3}\right) + \cot^{-1} \left(-\frac{1}{\sqrt{3}}\right) \), we will use some properties of inverse trigonometric functions.

### Step-by-step Solution:

1. **Using the property of sine inverse:**
\[ \sin^{-1}(-x) = -\sin^{-1}(x) \]
Thus,
\[ \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
We know that \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \). Therefore,
\[ \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4} \]

2. **Using the property of cosine inverse:**
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
Thus,
\[ \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) \]
We know that \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \). Therefore,
\[ \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

3. **Using the property of tangent inverse:**
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
Thus,
\[ -\tan^{-1}\left(-\sqrt{3}\right) = \tan^{-1}\left(\sqrt{3}\right) \]
We know that \( \tan^{-1}\left(\sqrt{3}\right) = \frac{\pi}{3} \). Therefore,
\[ -\tan^{-1}\left(-\sqrt{3}\right) = \frac{\pi}{3} \]

4. **Using the property of cotangent inverse:**
\[ \cot^{-1}(-x) = \pi - \cot^{-1}(x) \]
Thus,
\[ \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \pi - \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
We know that \( \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \). Therefore,
\[ \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]

5. **Combining all the results:**
Now we substitute back into the original expression:
\[ -\frac{\pi}{4} + \frac{2\pi}{3} + \frac{\pi}{3} + \frac{5\pi}{6} \]

To combine these fractions, we need a common denominator. The least common multiple of 4, 3, and 6 is 12.

- Convert each term:
\[ -\frac{\pi}{4} = -\frac{3\pi}{12} \]
\[ \frac{2\pi}{3} = \frac{8\pi}{12} \]
\[ \frac{\pi}{3} = \frac{4\pi}{12} \]
\[ \frac{5\pi}{6} = \frac{10\pi}{12} \]

- Now combine:
\[ -\frac{3\pi}{12} + \frac{8\pi}{12} + \frac{4\pi}{12} + \frac{10\pi}{12} = \frac{19\pi}{12} \]

### Final Answer:
The value of the expression is:
\[ \frac{19\pi}{12} \]