Question:
If $0<\mathrm{a}, \mathrm{b}<1$, and $\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\frac{\pi}{4}$, 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:
Correct Option: 1
Solution:
$\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\frac{\pi}{4} \quad 0<\mathrm{a}, \mathrm{b}<1$
$\Rightarrow \frac{a+b}{1-a b}=1$
$a+b=1-a b$
$(a+1)(b+1)=2$
Now $\left[a-\frac{a^{2}}{2}+\frac{a^{3}}{3}+\ldots\right]+\left[b-\frac{b^{2}}{2}+\frac{b^{3}}{3}+\ldots\right]$
$=\log _{e}(1+a)+\log _{e}(1+b)$
$\left(\because\right.$ expansion of $\left.\log _{\mathrm{e}}(1+\mathrm{x})\right)$
$=\log _{\mathrm{e}}[(1+\mathrm{a})(1+\mathrm{b})]$
$=\log _{\mathrm{e}} 2$