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

Options

A. `-pi/12`
B. `(11pi)/12`
C. `(5pi)/4`
D. `(23pi)/12`

Answer

To solve the expression \( \tan^{-1}(\sqrt{3}) + \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) + \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \), we will evaluate each term step by step.

### Step 1: Evaluate \( \tan^{-1}(\sqrt{3}) \)

The value of \( \tan^{-1}(\sqrt{3}) \) corresponds to the angle whose tangent is \( \sqrt{3} \). We know that:

\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

Thus,

\[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

### Step 2: Evaluate \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)

The value of \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \) corresponds to the angle whose cosine is \( -\frac{1}{\sqrt{2}} \). We know that:

\[ \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \]

Thus,

\[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \]

### Step 3: Evaluate \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \)

The value of \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \) corresponds to the angle whose secant is \( -\frac{2}{\sqrt{3}} \). We know that:

\[ \sec\left(\frac{5\pi}{6}\right) = -\frac{2}{\sqrt{3}} \]

Thus,

\[ \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) = \frac{5\pi}{6} \]

### Step 4: Combine the values

Now we can combine all the evaluated values:

\[ \tan^{-1}(\sqrt{3}) + \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) + \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) = \frac{\pi}{3} + \frac{3\pi}{4} + \frac{5\pi}{6} \]

### Step 5: Find a common denominator

The common denominator for \( 3, 4, \) and \( 6 \) is \( 12 \). We convert each term:

\[ \frac{\pi}{3} = \frac{4\pi}{12}, \quad \frac{3\pi}{4} = \frac{9\pi}{12}, \quad \frac{5\pi}{6} = \frac{10\pi}{12} \]

### Step 6: Add the fractions

Now we add the fractions:

\[ \frac{4\pi}{12} + \frac{9\pi}{12} + \frac{10\pi}{12} = \frac{4\pi + 9\pi + 10\pi}{12} = \frac{23\pi}{12} \]

### Final Answer

Thus, the value of \( \tan^{-1}(\sqrt{3}) + \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) + \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \) is:

\[ \frac{23\pi}{12} \]