Solve the following equations :
`tan^(-1) (x+1) +tan^(-1) (x-1) = tan^(-1) (8/31 ) , x gt 0 `

Answer

To solve the equation \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{8}{31}\right) \) where \( x > 0 \), we can use the identity for the sum of inverse tangents:

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

### Step-by-step Solution:

1. **Apply the Identity**:
We apply the identity to the left-hand side of the equation:
\[ \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{(x+1) + (x-1)}{1 - (x+1)(x-1)}\right) \]

2. **Simplify the Numerator**:
The numerator simplifies as follows:
\[ (x+1) + (x-1) = 2x \]

3. **Simplify the Denominator**:
The denominator simplifies as follows:
\[ 1 - (x+1)(x-1) = 1 - (x^2 - 1) = 1 - x^2 + 1 = 2 - x^2 \]

4. **Rewrite the Equation**:
Now we can rewrite the equation:
\[ \tan^{-1}\left(\frac{2x}{2 - x^2}\right) = \tan^{-1}\left(\frac{8}{31}\right) \]

5. **Remove the Inverse Tangent**:
Since the tangent function is one-to-one, we can equate the arguments:
\[ \frac{2x}{2 - x^2} = \frac{8}{31} \]

6. **Cross Multiply**:
Cross-multiplying gives:
\[ 2x \cdot 31 = 8(2 - x^2) \]
Simplifying this:
\[ 62x = 16 - 8x^2 \]

7. **Rearrange the Equation**:
Rearranging gives:
\[ 8x^2 + 62x - 16 = 0 \]

8. **Divide by 2**:
We can simplify the equation by dividing everything by 2:
\[ 4x^2 + 31x - 8 = 0 \]

9. **Factor the Quadratic**:
We need to factor the quadratic equation. We look for two numbers that multiply to \( 4 \times -8 = -32 \) and add to \( 31 \). The numbers are \( 32 \) and \( -1 \):
\[ 4x^2 + 32x - x - 8 = 0 \]
Grouping gives:
\[ 4x(x + 8) - 1(x + 8) = 0 \]
Factoring out \( (x + 8) \):
\[ (x + 8)(4x - 1) = 0 \]

10. **Find the Roots**:
Setting each factor to zero gives:
\[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \quad (\text{not valid since } x > 0) \]
\[ 4x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{4} \]

### Final Answer:
Thus, the solution to the equation is:
\[ \boxed{\frac{1}{4}} \]