Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If ∑r=0∞2 arccot(r2+r+42)=kπ\displaystyle\sum_{r=0}^{\infty}2\,\text{arccot}\left(\dfrac{r^{2}+r+4}{2}\right)=k\pi, then find the value of kk.
Solution
Answer: 1
Step 1: Convert to tan⁡−1\tan^{-1}:
Tr=2cot⁡−1r2+r+42=2tan⁡−12r2+r+4T_{r}=2\cot^{-1}\dfrac{r^{2}+r+4}{2}=2\tan^{-1}\dfrac{2}{r^{2}+r+4}
 Step 2: Manipulate the argument:
2r2+r+4=24+r(r+1)=r+12−r21+r2⋅r+12\dfrac{2}{r^{2}+r+4}=\dfrac{2}{4+r(r+1)}=\dfrac{\frac{r+1}{2}-\frac{r}{2}}{1+\frac{r}{2}\cdot\frac{r+1}{2}}
 Step 3: Apply difference formula:
Tr=2[tan⁡−1r+12−tan⁡−1r2]T_{r}=2\left[\tan^{-1}\dfrac{r+1}{2}-\tan^{-1}\dfrac{r}{2}\right]
 Step 4: Telescoping sum:
∑r=0nTr=2[tan⁡−1n+12−tan⁡−10]=2tan⁡−1n+12\sum_{r=0}^{n}T_{r}=2\left[\tan^{-1}\dfrac{n+1}{2}-\tan^{-1}0\right]=2\tan^{-1}\dfrac{n+1}{2}
 Step 5: Taking n→∞n\to\infty:
lim⁡n→∞2tan⁡−1n+12=2⋅π2=π=kπ\lim_{n\to\infty}2\tan^{-1}\dfrac{n+1}{2}=2\cdot\dfrac{\pi}{2}=\pi=k\pi
 Step 6: So k=1k=1. Answer: 11
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