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Question:

If $0<a, b<1$, and $\tan ^{-1} a+\tan ^{-1} b=\frac{\pi}{4}$, then the value of $(a+b)-\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{3}+b^{3}}{3}\right)-\left(\frac{a^{4}+b^{4}}{4}\right)+\ldots$ is :

  1. (1) $\log _{e} 2$

  2. (2) $\log _{e}\left(\frac{e}{2}\right)$

  3. (3) $e$

  4. (4) $e^{2}-1$


Correct Option: 1,

Solution:

$\tan ^{-1}\left(\frac{a+b}{1-a b}\right)=\frac{\pi}{4} \Rightarrow a+b=1-a b \Rightarrow(1+a)(1+b)=2$

Now, $(a+b)-\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{3}+b^{3}}{3}\right) \cdots \infty$

$=\left(a-\frac{a^{2}}{2}+\frac{a^{3}}{3} \cdots\right)+\left(b-\frac{b^{2}}{2}+\frac{b^{3}}{3} \cdots\right)$

$\log _{e}(1+a)+\log _{e}(1+b)=\log _{e}(1+a)(1+b)=\log _{0} 2$

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