`tan^-1a+tan^-1b=pi/4 .` Find the value of `a+b-(a^2+b^2)/2+(a^3+b^3)/3-(a^4+b^4)/5+ . . . `

Answer

To solve the problem, we start with the equation given:

\[ \tan^{-1} a + \tan^{-1} b = \frac{\pi}{4} \]

Using the identity for the sum of inverse tangents, we have:

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

Setting this equal to \(\frac{\pi}{4}\), we can take the tangent of both sides:

\[ \frac{a + b}{1 - ab} = 1 \]

Cross-multiplying gives:

\[ a + b = 1 - ab \]

Rearranging this equation, we find:

\[ a + b + ab = 1 \]

Next, we need to find the value of the series:

\[ S = a + b - \frac{a^2 + b^2}{2} + \frac{a^3 + b^3}{3} - \frac{a^4 + b^4}{5} + \ldots \]

We can express \(a^2 + b^2\) and \(a^3 + b^3\) in terms of \(a + b\) and \(ab\):

1. **Finding \(a^2 + b^2\)**:
\[ a^2 + b^2 = (a + b)^2 - 2ab = (1 - ab)^2 - 2ab = 1 - 2ab + a^2b^2 - 2ab = 1 - 4ab + a^2b^2 \]

2. **Finding \(a^3 + b^3\)**:
\[ a^3 + b^3 = (a + b)(a^2 + b^2 - ab) = (1 - ab)(1 - 4ab + a^2b^2 - ab) \]

3. **Finding \(a^4 + b^4\)**:
\[ a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \]

Now, we can rewrite the series \(S\) using these identities.

The series can be recognized as the Taylor series expansion for \(\ln(1+x)\):

\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

Thus, we can express \(S\) as:

\[ S = \ln(1+a) + \ln(1+b) = \ln((1+a)(1+b)) \]

Using the relationship \(a + b + ab = 1\), we can write:

\[ (1 + a)(1 + b) = 1 + (a + b) + ab = 1 + 1 = 2 \]

Therefore, we have:

\[ S = \ln(2) \]

### Final Answer:

\[ \boxed{\ln(2)} \]