Inverse Trigonometry
Infinite telescoping series of cot⁻¹
MMTS_Full_Test_05
Grade 12
Question:
If $a=2$, then the sum of the infinite series $\cot^{-1}(2a^{-1}+a)+\cot^{-1}(2a^{-1}+3a)+\cot^{-1}(2a^{-1}+6a)+\cot^{-1}(2a^{-1}+10a)+\cdots$ is
(A) $\dfrac{\pi}{4}$
(B) $\dfrac{\pi}{2}$
(C) $\dfrac{\pi}{3}$
(D) $\dfrac{\pi}{6}$
Step-by-Step Solution
Key Concept: With $a=2$: terms are $\cot^{-1}(1+n(n+1))=\tan^{-1}\frac{1}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}(n)$. Telescopes.
Sum $=\pi/4$.
Correct Answer: (A) $\dfrac{\pi}{4}$