The value of `"tan"^(-1)(1)/(2)+"tan"^(-1)(1)/(3)+"tan"^(-1)(7)/(8)` is

Options

A. `"tan"^(-1)(7)/(8)`
B. `cot^(-1)15`
C. `tan^(-1)15`
D. `"tan"^(-1)(25)/(24)`

Explanation

To solve the expression \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{7}{8}\right) \), we will use the formula for the sum of inverse tangents:

\[ \tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A + B}{1 - AB}\right) \]

### Step 1: Combine the first two terms
Let \( A = \frac{1}{2} \) and \( B = \frac{1}{3} \).

Using the formula:

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

### Step 2: Calculate the numerator and denominator
1. **Numerator**:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

2. **Denominator**:
\[ 1 - \frac{1}{2} \cdot \frac{1}{3} = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Combine the results
Now substituting back into the formula:

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

### Step 4: Simplify
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:

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

### Step 5: Add the third term
Now we need to add \( \tan^{-1}\left(\frac{7}{8}\right) \):

\[ \tan^{-1}(1) + \tan^{-1}\left(\frac{7}{8}\right) = \tan^{-1}\left(\frac{1 + \frac{7}{8}}{1 - 1 \cdot \frac{7}{8}}\right) \]

### Step 6: Calculate the new numerator and denominator
1. **Numerator**:
\[ 1 + \frac{7}{8} = \frac{8}{8} + \frac{7}{8} = \frac{15}{8} \]

2. **Denominator**:
\[ 1 - 1 \cdot \frac{7}{8} = 1 - \frac{7}{8} = \frac{1}{8} \]

### Step 7: Combine the results again
Now substituting back:

\[ \tan^{-1}\left(\frac{\frac{15}{8}}{\frac{1}{8}}\right) = \tan^{-1}(15) \]

### Final Result
Thus, the value of the expression is:

\[ \tan^{-1}(15) \]