JEE MainMathematics

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Telescoping Sum of Inverse Tangents

Q: Does the identity tan^(-1)(x) - tan^(-1)(y) = tan^(-1)((x-y)/(1+xy)) hold without extra π shifts here?

Yes, because both x = k+1 and y = k are positive natural numbers, meaning xy > 0 > -1, where the principal identity holds unconditionally.

Let a_1 = 1, a_2, a_3, a_4, … be consecutive natural numbers. Then tan^(-1)(1/(1 + a_1a_2)) + tan^(-1)(1/(1 + a_2a_3)) + … + tan^(-1)(1/(1 + a_2021a_2022)) is equal to

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Acot^(-1)(2022) - π/4Bπ/4 - cot^(-1)(2022)Ctan^(-1)(2022) - π/4Dπ/4 - tan^(-1)(2022)

Step-by-step solutionView

Correct answer

The sum telescopes via the identity $\tan^{- 1} (\frac{x - y}{1 + x y}) = \tan^{- 1} x - \tan^{- 1} y$ , evaluating to $\tan^{- 1} (2022) - \frac{π}{4}$ .

Option analysis

Why each option works or fails

A · cot^(-1)(2022) - π/4

Confusing the functions $\tan^{- 1} (x)$ and $\cot^{- 1} (x)$ , incorrectly replacing $\tan^{- 1} (2022)$ with $\cot^{- 1} (2022)$ . Recall that $\tan^{- 1} (x) = \cot^{- 1} (1 / x)$ for $x > 0$ , so $\tan^{- 1} (2022)$ is not equal to $\cot^{- 1} (2022)$ .

B · π/4 - cot^(-1)(2022)

Reversing the telescoping order and concurrently substituting $\cot^{- 1}$ for $\tan^{- 1}$ . Keep track of the signs during subtraction: the higher term $a_{k + 1}$ carries a positive sign, giving $\tan^{- 1} (a_{2022}) - \tan^{- 1} (a_{1})$ .

C · tan^(-1)(2022) - π/4

This is the correct option. Writing the general term as $\tan^{- 1} (k + 1) - \tan^{- 1} (k)$ creates a telescoping sum whose first and last terms are $- \tan^{- 1} (1) = - π / 4$ and $\tan^{- 1} (2022)$ .

D · π/4 - tan^(-1)(2022)

Subtracting in reverse order, writing the general term as $\tan^{- 1} (a_{k}) - \tan^{- 1} (a_{k + 1})$ instead of $\tan^{- 1} (a_{k + 1}) - \tan^{- 1} (a_{k})$ . Because $a_{k + 1} - a_{k} = 1 > 0$ , the numerator $1$ matches $a_{k + 1} - a_{k}$ , which yields $\tan^{- 1} (a_{k + 1}) - \tan^{- 1} (a_{k})$ .

Reviewed route

Solution

StepWorking

01Given

$a_{1} = 1, a_{2} = 2, \dots, a_{n} = n$ are consecutive natural numbers, and the sum to evaluate is $S = \sum_{k = 1}^{2021} \tan^{- 1} (\frac{1}{1 + a_{k} a_{k + 1}})$ .

02Goal

Evaluate the summation $\sum_{k = 1}^{2021} \tan^{- 1} (\frac{1}{1 + k (k + 1)})$ in terms of inverse trigonometric functions.

03Approach

Rewrite the numerator $1$ as $(k + 1) - k$ in each term so that $\tan^{- 1} (\frac{(k + 1) - k}{1 + k (k + 1)}) = \tan^{- 1} (k + 1) - \tan^{- 1} (k)$ , yielding a telescoping sum.

04Execute

For the general term $T_{k} = \tan^{- 1} (\frac{(k + 1) - k}{1 + (k + 1) k}) = \tan^{- 1} (k + 1) - \tan^{- 1} (k)$ . Summing from $k = 1$ to $2021$ gives $S = (\tan^{- 1} 2 - \tan^{- 1} 1) + (\tan^{- 1} 3 - \tan^{- 1} 2) + \dots + (\tan^{- 1} 2022 - \tan^{- 1} 2021) = \tan^{- 1} (2022) - \tan^{- 1} (1) = \tan^{- 1} (2022) - \frac{π}{4}$ .

✓Verify

Test $n = 1$ : sum is $\tan^{- 1} (1 / (1 + 1 \cdot 2)) = \tan^{- 1} (1 / 3)$ . From the formula with upper limit $a_{2} = 2$ : $\tan^{- 1} (2) - \tan^{- 1} (1) = \tan^{- 1} ((2 - 1) / (1 + 2)) = \tan^{- 1} (1 / 3)$ , which perfectly matches.

Hints that build this answer step by step

Since $a_{k} = k$ , how can the term $\tan^{- 1} (\frac{1}{1 + a_{k} a_{k + 1}})$ be decomposed using an inverse trigonometric identity?
$\tan^{- 1} (k + 1) - \tan^{- 1} (k)$
Summing the decomposed terms $\sum_{k = 1}^{2021} (\tan^{- 1} (k + 1) - \tan^{- 1} (k))$ , which terms survive after cancellation?
$\tan^{- 1} (2022) - \tan^{- 1} (1)$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Does the identity

\tan^{- 1} (x) - \tan^{- 1} (y) = \tan^{- 1} (\frac{x - y}{1 + x y})

hold without extra

π

shifts here?

Yes, because both $x = k + 1$ and $y = k$ are positive natural numbers, meaning $x y > 0 > - 1$ , where the principal identity holds unconditionally.

Answer

The sum telescopes via the identity $\tan^{- 1} (\frac{x - y}{1 + x y}) = \tan^{- 1} x - \tan^{- 1} y$ , evaluating to $\tan^{- 1} (2022) - \frac{π}{4}$ .

Why each option works or fails

A: cot^(-1)(2022) - π/4 - Confusing the functions $\tan^{- 1} (x)$ and $\cot^{- 1} (x)$ , incorrectly replacing $\tan^{- 1} (2022)$ with $\cot^{- 1} (2022)$ . Recall that $\tan^{- 1} (x) = \cot^{- 1} (1 / x)$ for $x > 0$ , so $\tan^{- 1} (2022)$ is not equal to $\cot^{- 1} (2022)$ .
B: π/4 - cot^(-1)(2022) - Reversing the telescoping order and concurrently substituting $\cot^{- 1}$ for $\tan^{- 1}$ . Keep track of the signs during subtraction: the higher term $a_{k + 1}$ carries a positive sign, giving $\tan^{- 1} (a_{2022}) - \tan^{- 1} (a_{1})$ .
C · correct: tan^(-1)(2022) - π/4 - This is the correct option. Writing the general term as $\tan^{- 1} (k + 1) - \tan^{- 1} (k)$ creates a telescoping sum whose first and last terms are $- \tan^{- 1} (1) = - π / 4$ and $\tan^{- 1} (2022)$ .
D: π/4 - tan^(-1)(2022) - Subtracting in reverse order, writing the general term as $\tan^{- 1} (a_{k}) - \tan^{- 1} (a_{k + 1})$ instead of $\tan^{- 1} (a_{k + 1}) - \tan^{- 1} (a_{k})$ . Because $a_{k + 1} - a_{k} = 1 > 0$ , the numerator $1$ matches $a_{k + 1} - a_{k}$ , which yields $\tan^{- 1} (a_{k + 1}) - \tan^{- 1} (a_{k})$ .

Step-by-step solution

Given: $a_{1} = 1, a_{2} = 2, \dots, a_{n} = n$ are consecutive natural numbers, and the sum to evaluate is $S = \sum_{k = 1}^{2021} \tan^{- 1} (\frac{1}{1 + a_{k} a_{k + 1}})$ .
Goal: Evaluate the summation $\sum_{k = 1}^{2021} \tan^{- 1} (\frac{1}{1 + k (k + 1)})$ in terms of inverse trigonometric functions.
Approach: Rewrite the numerator $1$ as $(k + 1) - k$ in each term so that $\tan^{- 1} (\frac{(k + 1) - k}{1 + k (k + 1)}) = \tan^{- 1} (k + 1) - \tan^{- 1} (k)$ , yielding a telescoping sum.
Execute: For the general term $T_{k} = \tan^{- 1} (\frac{(k + 1) - k}{1 + (k + 1) k}) = \tan^{- 1} (k + 1) - \tan^{- 1} (k)$ . Summing from $k = 1$ to $2021$ gives $S = (\tan^{- 1} 2 - \tan^{- 1} 1) + (\tan^{- 1} 3 - \tan^{- 1} 2) + \dots + (\tan^{- 1} 2022 - \tan^{- 1} 2021) = \tan^{- 1} (2022) - \tan^{- 1} (1) = \tan^{- 1} (2022) - \frac{π}{4}$ .
Verify: Test $n = 1$ : sum is $\tan^{- 1} (1 / (1 + 1 \cdot 2)) = \tan^{- 1} (1 / 3)$ . From the formula with upper limit $a_{2} = 2$ : $\tan^{- 1} (2) - \tan^{- 1} (1) = \tan^{- 1} ((2 - 1) / (1 + 2)) = \tan^{- 1} (1 / 3)$ , which perfectly matches.

Telescoping Sum of Inverse Tangents

What feels right?

The sum telescopes via the identity $\tan^{- 1} (\frac{x - y}{1 + x y}) = \tan^{- 1} x - \tan^{- 1} y$ , evaluating to $\tan^{- 1} (2022) - \frac{π}{4}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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01Given

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