text: # If $0, and $\tan ^{-1}a + \tan ^{-1}b =\frac{\pi}{4}$, then the value of  $(a+b)-\left ( \frac{a^2+b^2}{2} \right )+\left ( \frac{a^3+b^3}{3} \right )-\left ( \frac{a^4+b^4}{4} \right )+.........$ is: Option: 1 e Option: 3 $\log _e\left (\frac{e}{2} \right )$ Option: 5 $\log _e 2$ Option: 7 $e^2-1$
  
  $\\\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\frac{\pi}{4} \quad\quad 0<\mathrm{a}, \mathrm{b}<1 \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}=1 \\ \mathrm{a}+\mathrm{b}=1-\mathrm{ab} \\ (\mathrm{a}+1)(\mathrm{b}+1)=2$
  
  $\\\text { Now }\left[a-\frac{a^{2}}{2}+\frac{a^{3}}{3}+\ldots\right]+\left[b-\frac{b^{2}}{2}+\frac{b^{3}}{3}+\ldots\right] \\ =\log _{e}(1+a)+\log _{e}(1+b) \\ \left(\because \text { expansion of } \log _{e}(1+x)\right) \\ =\log _{e}[(1+a)(1+b)] \\ =\log _{e} 2$
  
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