Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions: Coprime Find Value

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

Question
If cot⁡−1(4+24)+cot⁡−1(4+64)+cot⁡−1(4+124)+…∞=tan⁡−1(ab)\cot^{-1}\left(4+\dfrac{2}{4}\right)+\cot^{-1}\left(4+\dfrac{6}{4}\right)+\cot^{-1}\left(4+\dfrac{12}{4}\right)+\ldots\infty=\tan^{-1}\left(\dfrac{a}{b}\right) where aa and bb are coprime, then find the value of (a3+b3)(a^{3}+b^{3}).
Solution
Answer: 65
Step 1: Observe the pattern: 2,6,12,…2,6,12,\ldots has general term r(r+1)r(r+1). So the rr-th term is:
Tr=cot⁡−1(4+r(r+1)4)=cot⁡−1(16+r(r+1)4)=tan⁡−1416+r(r+1)T_{r}=\cot^{-1}\left(4+\dfrac{r(r+1)}{4}\right)=\cot^{-1}\left(\dfrac{16+r(r+1)}{4}\right)=\tan^{-1}\dfrac{4}{16+r(r+1)}
 Step 2: Manipulate the argument:
416+r(r+1)=r+14−r41+r4⋅r+14\dfrac{4}{16+r(r+1)}=\dfrac{\frac{r+1}{4}-\frac{r}{4}}{1+\frac{r}{4}\cdot\frac{r+1}{4}}
 Step 3: So:
Tr=tan⁡−1r+14−tan⁡−1r4T_{r}=\tan^{-1}\dfrac{r+1}{4}-\tan^{-1}\dfrac{r}{4}
 Step 4: Telescoping sum:
Sn=∑r=1nTr=tan⁡−1n+14−tan⁡−114S_{n}=\sum_{r=1}^{n}T_{r}=\tan^{-1}\dfrac{n+1}{4}-\tan^{-1}\dfrac{1}{4}
 Step 5: Taking n→∞n\to\infty:
S∞=π2−tan⁡−114=cot⁡−114=tan⁡−14=tan⁡−1abS_{\infty}=\dfrac{\pi}{2}-\tan^{-1}\dfrac{1}{4}=\cot^{-1}\dfrac{1}{4}=\tan^{-1}4=\tan^{-1}\dfrac{a}{b}
 Step 6: So a=4a=4 and b=1b=1. Hence:
a3+b3=64+1=65a^{3}+b^{3}=64+1=65
 Answer: 6565
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