Write the value of `tan^(-1)2+tan^(-1)3`.

Answer

To solve the expression \( \tan^{-1}(2) + \tan^{-1}(3) \), we can use the formula for the sum of two inverse tangents:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]

### Step 1: Identify \( x \) and \( y \)
Here, we have \( x = 2 \) and \( y = 3 \).

### Step 2: Substitute \( x \) and \( y \) into the formula
Substituting the values into the formula gives us:

\[ \tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}\left(\frac{2 + 3}{1 - (2 \cdot 3)}\right) \]

### Step 3: Simplify the expression
Now, we simplify the fraction:

\[ = \tan^{-1}\left(\frac{5}{1 - 6}\right) = \tan^{-1}\left(\frac{5}{-5}\right) = \tan^{-1}(-1) \]

### Step 4: Evaluate \( \tan^{-1}(-1) \)
The value of \( \tan^{-1}(-1) \) corresponds to an angle where the tangent is -1. This occurs at:

\[ \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Final Answer
Thus, the value of \( \tan^{-1}(2) + \tan^{-1}(3) \) is:

\[ -\frac{\pi}{4} \]

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