If `tan^(-1)3+ tan^(-1)x = tan^(-1)8`, then: `x=`

Options

A. `5`
B. `1/5`
C. `5/14`
D. `14/5`

Answer

To solve the equation \( \tan^{-1}(3) + \tan^{-1}(x) = \tan^{-1}(8) \), we can use the formula for the sum of two inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

### Step-by-Step Solution:

1. **Apply the formula for the sum of inverse tangents**:
\[ \tan^{-1}(3) + \tan^{-1}(x) = \tan^{-1}\left(\frac{3 + x}{1 - 3x}\right) \]
Thus, we have:
\[ \tan^{-1}\left(\frac{3 + x}{1 - 3x}\right) = \tan^{-1}(8) \]

2. **Eliminate the inverse tangent**:
Since \( \tan^{-1}(A) = \tan^{-1}(B) \) implies \( A = B \), we can set the arguments equal to each other:
\[ \frac{3 + x}{1 - 3x} = 8 \]

3. **Cross-multiply to eliminate the fraction**:
\[ 3 + x = 8(1 - 3x) \]
Expanding the right side gives:
\[ 3 + x = 8 - 24x \]

4. **Rearrange the equation**:
Move all terms involving \( x \) to one side and constant terms to the other side:
\[ x + 24x = 8 - 3 \]
Simplifying this results in:
\[ 25x = 5 \]

5. **Solve for \( x \)**:
Divide both sides by 25:
\[ x = \frac{5}{25} = \frac{1}{5} \]

### Final Answer:
\[ x = \frac{1}{5} \]