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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 )+......... is:
Option: 1 e
Option: 3 \log _e\left (\frac{e}{2} \right )
Option: 5 \log _e 2
Option: 7 e^2-1

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\\\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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himanshu.meshram

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