JEE PYQ: Inverse Trigonometric Functions Question 10
Question 10 - 2021 (26 Feb Shift 2)
If $0 < a, b < 1$ and $\tan^{-1}a + \tan^{-1}b = \frac{\pi}{4}$, then the value of $(a+b) - \left(\frac{a^2+b^2}{2}\right) + \left(\frac{a^3+b^3}{3}\right) - \left(\frac{a^4+b^4}{4}\right) + \ldots$ is:
(1) $\log_e 2$
(2) $\log_e\left(\frac{e}{2}\right)$
(3) $e$
(4) $e^2 - 1$
Show Answer
Answer: (1)
Solution
$\tan^{-1}a + \tan^{-1}b = \frac{\pi}{4} \Rightarrow \frac{a+b}{1-ab} = 1 \Rightarrow a + b = 1 - ab$. The series is $\left(a - \frac{a^2}{2} + \frac{a^3}{3} - \ldots\right) + \left(b - \frac{b^2}{2} + \frac{b^3}{3} - \ldots\right) = \log_e(1+a) + \log_e(1+b) = \log_e((1+a)(1+b)) = \log_e(1 + a + b + ab)$. Since $a + b + ab = 1$: $\log_e 2$.
BETA
