Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

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

Question
If lim⁡n→∞∑k=1ncot⁡−1(1+k+k2)=cot⁡−1(α)+cot⁡−1(β)\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\cot^{-1}(1+k+k^{2})=\cot^{-1}(\alpha)+\cot^{-1}(\beta), where α,β\alpha,\beta are prime numbers, then find (α+β)(\alpha+\beta).
Solution
Answer: 5
Step 1: Convert to tan⁡−1\tan^{-1}:
cot⁡−1(1+k+k2)=tan⁡−111+k+k2=tan⁡−1(k+1)−k1+(k+1)k\cot^{-1}(1+k+k^{2})=\tan^{-1}\dfrac{1}{1+k+k^{2}}=\tan^{-1}\dfrac{(k+1)-k}{1+(k+1)k}
 Step 2: Apply the difference formula:
=tan⁡−1(k+1)−tan⁡−1k=\tan^{-1}(k+1)-\tan^{-1}k
 Step 3: Telescoping sum:
∑k=1n[tan⁡−1(k+1)−tan⁡−1k]=tan⁡−1(n+1)−tan⁡−11\sum_{k=1}^{n}\left[\tan^{-1}(k+1)-\tan^{-1}k\right]=\tan^{-1}(n+1)-\tan^{-1}1
 Step 4: Taking n→∞n\to\infty:
lim⁡n→∞tan⁡−1(n+1)−tan⁡−11=π2−π4=π4\lim_{n\to\infty}\tan^{-1}(n+1)-\tan^{-1}1=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}
 Step 5: Express π4\dfrac{\pi}{4} as a sum of two cot⁡−1\cot^{-1} of primes:
π4=tan⁡−11=tan⁡−112+tan⁡−113=cot⁡−12+cot⁡−13\dfrac{\pi}{4}=\tan^{-1}1=\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}=\cot^{-1}2+\cot^{-1}3
 Verification: tan⁡−112+tan⁡−113=tan⁡−11/2+1/31−1/6=tan⁡−15/65/6=tan⁡−11=π4\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}=\tan^{-1}\dfrac{1/2+1/3}{1-1/6}=\tan^{-1}\dfrac{5/6}{5/6}=\tan^{-1}1=\dfrac{\pi}{4}. ✓ Step 6: So α=2,β=3\alpha=2,\beta=3 (both prime), giving α+β=5\alpha+\beta=5. Answer: 55
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