JEE Mathematics 2023 Question with Solution

Given: $a_1 = 1$ and $a_1, a_2, a_3, \dots$ are consecutive natural numbers.

Find:

$$\tan^{-1} \left( \frac{1}{1 + a_1 a_2} \right) + \tan^{-1} \left( \frac{1}{1 + a_2 a_3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + a_{2021}a_{2022}} \right)$$

Since the terms are consecutive natural numbers,

$$a_2-a_1=a_3-a_2=\dots=a_{2022}-a_{2021}=1$$

So each term can be written as

$$\tan^{-1}\left(\frac{1+a_i a_{i+1}}{a_{i+1}-a_i}\right)=\tan^{-1}(a_{i+1})-\tan^{-1}(a_i)$$

using the tangent subtraction pattern shown in the solution. Hence,

$$\tan^{-1} \left( \frac{1}{1 + a_1 a_2} \right)+\tan^{-1} \left( \frac{1}{1 + a_2 a_3} \right)+\dots+\tan^{-1} \left( \frac{1}{1 + a_{2021}a_{2022}} \right)$$

becomes

$$[\tan^{-1}(a_2)-\tan^{-1}(a_1)] + [\tan^{-1}(a_3)-\tan^{-1}(a_2)] + \dots + [\tan^{-1}(a_{2022})-\tan^{-1}(a_{2021})]$$

The series telescopes, so all intermediate terms cancel:

$$[\tan^{-1}(a_2)-\tan^{-1}(a_1)] + [\tan^{-1}(a_3)-\tan^{-1}(a_2)] + \dots + [\tan^{-1}(a_{2022})-\tan^{-1}(a_{2021})]$$ $$= \tan^{-1}(a_{2022})-\tan^{-1}(a_1)$$

Because $a_1=1$ and the numbers are consecutive natural numbers,

$$a_{2022}=2022$$

Therefore,

$$\tan^{-1}(a_{2022})-\tan^{-1}(a_1)=\tan^{-1}(2022)-\tan^{-1}(1)$$ $$= \tan^{-1}(2022)-\frac{\pi}{4}$$

Using $\tan^{-1}(x)=\frac{\pi}{2}-\cot^{-1}(x)$,

$$\tan^{-1}(2022)-\frac{\pi}{4}=\frac{\pi}{4}-\cot^{-1}(2022)$$

the solution marks B as correct, but this final expression matches Option A. Therefore, the solution contains an option-label discrepancy, and the defensible correct option is A.