Question

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

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

Answer 1

The correct option is A

${\log }_e\left(2\right)$

Explanation for the correct option.

Find the value of the expression.

It is known that ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$.

So the given equation ${\tan }^{-1}a+{\tan }^{-1}b=\frac{\mathrm{\pi }}{4}$ can be written as:

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

Now, using the expansion ${\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$ the expression $\left(a+b\right)-\frac{a^2+b^2}{2}+\frac{a^3+b^3}{3}-\frac{a^4+b^4}{4}+...$ can be evaluated as:

$\begin{array}{rcl}\left(a+b\right)-\frac{a^2+b^2}{2}+\frac{a^3+b^3}{3}-\frac{a^4+b^4}{4}+...& =& \left(a-\frac{a^2}{2}+\frac{a^3}{3}-\frac{a^4}{4}+...\right)+\left(b-\frac{b^2}{2}+\frac{b^3}{3}-\frac{b^4}{4}+...\right)\\ & =& {\log }_e\left(1+a\right)+{\log }_e\left(1+b\right)\\ & =& {\log }_e\left(\left(1+a\right)\left(1+b\right)\right)\left[{\log }_a\left(m\right)+{\log }_a\left(n\right)={\log }_a\left(mn\right)\right]\\ & =& {\log }_e\left(2\right)\left[\left(1+a\right)\left(1+b\right)=2\right]\end{array}$

So, the value of the expression is ${\log }_e\left(2\right)$.

Hence, the correct option is A.