Question

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

• A. ${\log }_{e}2$
• B. ${\log }_e\left(\frac{e}{2}\right)$
• C. $e$
• D. $e^2-1$

Answer 1

The correct option is A ${\log }_{e}2$

${\tan }^{-1}\left(\frac{a+b}{1-ab}\right)=\frac{\pi }{4}\Rightarrow a+b=1-ab\Rightarrow (1+a)(1+b)=2$
Now,
$(a+b)-\left(\frac{a^2+b^2}{2}\right)+\left(\frac{a^3+b^3}{3}\right)+\cdots \infty$
$=(a-\frac{a^2}{2}+\frac{a^3}{3}-\cdots )+(b-\frac{b^2}{2}+\frac{b^3}{3}-\dots )$
$={\log }_e(1+a)+{\log }_e(1+b)={\log }_e(1+a)(1+b)={\log }_{e}2$