Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions: Prime Numbers Find

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

Question

\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\cot^{-1}(1+k+k^{2})=\cot^{-1}(\alpha)+\cot^{-1}(\beta)

, where

\alpha,\beta

are prime numbers, then find

(\alpha+\beta)

Solution

Answer: 5

Step 1: Convert to

\tan^{-1}

\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^{-1}k

Step 3: Telescoping sum:

\sum_{k=1}^{n}\left[\tan^{-1}(k+1)-\tan^{-1}k\right]=\tan^{-1}(n+1)-\tan^{-1}1

Step 4: Taking

n\to\infty

\lim_{n\to\infty}\tan^{-1}(n+1)-\tan^{-1}1=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}

Step 5: Express

\dfrac{\pi}{4}

as a sum of two

\cot^{-1}

of primes:

\dfrac{\pi}{4}=\tan^{-1}1=\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}=\cot^{-1}2+\cot^{-1}3

Verification:

\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

\alpha=2,\beta=3

(both prime), giving

\alpha+\beta=5

. Answer:

5

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