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

Options

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

Explanation

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

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

provided \( ab < 1 \). In this case, we need to rearrange the equation to isolate \( \tan^{-1}(x) \).

### Step 1: Rearranging the equation
Starting from the original equation:

\[ \tan^{-1}(x) = \tan^{-1}(8) - \tan^{-1}(3) \]

### Step 2: Applying the formula
Now, we can apply the formula for the difference of inverse tangents:

\[ \tan^{-1}(8) - \tan^{-1}(3) = \tan^{-1}\left(\frac{8 - 3}{1 + 8 \cdot 3}\right) \]

### Step 3: Calculating the values
Now, substitute \( a = 8 \) and \( b = 3 \):

\[ \tan^{-1}\left(\frac{8 - 3}{1 + 8 \cdot 3}\right) = \tan^{-1}\left(\frac{5}{1 + 24}\right) = \tan^{-1}\left(\frac{5}{25}\right) \]

### Step 4: Simplifying the fraction
This simplifies to:

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

### Step 5: Equating to find \( x \)
Since \( \tan^{-1}(x) = \tan^{-1}\left(\frac{1}{5}\right) \), we can conclude that:

\[ x = \frac{1}{5} \]

Thus, the final answer is:

\[ \boxed{\frac{1}{5}} \]