Find `tan^(-1)(3)+cot^(-1)(-1/3)+sec^(-1)2`.

Answer

To solve the expression \( \tan^{-1}(3) + \cot^{-1}(-\frac{1}{3}) + \sec^{-1}(2) \), we will break it down step by step.

### Step 1: Rewrite \( \cot^{-1}(-\frac{1}{3}) \)
We know that:
\[ \cot^{-1}(x) = \pi - \tan^{-1}(x) \quad \text{for } x < 0 \]
Since \(-\frac{1}{3} < 0\), we can write:
\[ \cot^{-1}(-\frac{1}{3}) = \pi - \tan^{-1}(-\frac{1}{3}) \]

### Step 2: Simplify \( \tan^{-1}(-\frac{1}{3}) \)
Using the property of the inverse tangent function:
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
we have:
\[ \tan^{-1}(-\frac{1}{3}) = -\tan^{-1}(\frac{1}{3}) \]
Thus,
\[ \cot^{-1}(-\frac{1}{3}) = \pi + \tan^{-1}(\frac{1}{3}) \]

### Step 3: Substitute back into the expression
Now we substitute this back into the original expression:
\[ \tan^{-1}(3) + \cot^{-1}(-\frac{1}{3}) + \sec^{-1}(2) = \tan^{-1}(3) + \left(\pi + \tan^{-1}(\frac{1}{3})\right) + \sec^{-1}(2) \]
This simplifies to:
\[ \tan^{-1}(3) + \tan^{-1}(\frac{1}{3}) + \pi + \sec^{-1}(2) \]

### Step 4: Use the property of \( \tan^{-1} \)
We know that:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad \text{if } xy < 1 \]
In our case, \( x = 3 \) and \( y = \frac{1}{3} \):
\[ xy = 3 \cdot \frac{1}{3} = 1 \quad \text{(not valid for this property)} \]
Instead, we can directly evaluate:
\[ \tan^{-1}(3) + \tan^{-1}(\frac{1}{3}) = \frac{\pi}{4} \quad \text{(since } \tan(\frac{\pi}{4}) = 1\text{)} \]

### Step 5: Substitute back into the expression
Now we substitute this back:
\[ \frac{\pi}{4} + \pi + \sec^{-1}(2) \]

### Step 6: Evaluate \( \sec^{-1}(2) \)
We know that:
\[ \sec^{-1}(x) = \frac{\pi}{3} \quad \text{(since } \sec(\frac{\pi}{3}) = 2\text{)} \]

### Step 7: Combine all parts
Now we can combine everything:
\[ \frac{\pi}{4} + \pi + \frac{\pi}{3} \]
To add these, we need a common denominator. The least common multiple of 4, 1, and 3 is 12:
\[ \frac{\pi}{4} = \frac{3\pi}{12}, \quad \pi = \frac{12\pi}{12}, \quad \frac{\pi}{3} = \frac{4\pi}{12} \]
Adding these gives:
\[ \frac{3\pi}{12} + \frac{12\pi}{12} + \frac{4\pi}{12} = \frac{19\pi}{12} \]

### Final Answer
Thus, the final answer is:
\[ \frac{19\pi}{12} \]