Animated Solution for Mathematics - Inverse Trigonometric Functions: If $0<a,b<1$, and ${\tan }^{-1}a+{\tan }^{-1}b=\frac{\pi }{4}$, then the value of $(a+b)-\frac{(a^2+b^2)}{2}+\frac{(a^3+b^3)}{3}-\frac{(a^4+b^4)}{4}+\dots$ is :

We begin with the condition ${\tan }^{-1}a+{\tan }^{-1}b=\frac{\pi }{4}$. This is our anchor.

We utilize the standard identity:

By applying this, we transform our equation into:

Taking the tangent of both sides, we obtain:

This is the moment where the trigonometry vanishes, leaving behind a pure algebraic skeleton: $a+b=1-ab$, or $a+b+ab=1$. We have successfully stripped away the complexity of the inverse trigonometric functions to reveal a simple relationship between $a$ and $b$.

Now, consider the relation $a+b+ab=1$. This is a classic setup that invites factorization.

If we add $1$ to both sides, we get:

Factoring the left side, we see $(1+a)+b(1+a)=2$, which simplifies to:

This is the key that unlocks the entire problem. We have taken a seemingly unrelated trigonometric condition and distilled it into a simple product, demonstrating the essence of JEE Advanced problem-solving: finding the hidden simplicity within complexity.

Now, consider the target series:

If we group the terms by variable, we see:

Recognizing this as the Taylor expansion of $\ln (1+x)$, we can write:

Using the logarithmic property $\ln M+\ln N=\ln (MN)$, we get:

Substituting our earlier result $(1+a)(1+b)=2$, we arrive at the final result: $S=\ln (2)$