Does the identity hold without extra shifts here?

Yes, because both and are positive natural numbers, meaning , where the principal identity holds unconditionally.

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Trigonometry: JEE Main Mathematics Question with Solution

Let

\text{a}_1 = 1, \text{a}_2, \text{a}_3, \text{a}_4, \dots

be consecutive natural numbers. Then

\tan^{-1}\left(\frac{1}{1 + \text{a}_1\text{a}_2}\right) + \tan^{-1}\left(\frac{1}{1 + \text{a}_2\text{a}_3}\right) + \dots + \tan^{-1}\left(\frac{1}{1 + \text{a}_{2021}\text{a}_{2022}}\right)

is equal to

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

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

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

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

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

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\cot^{-1}(2022) - \frac{\pi}{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 ·

\frac{\pi}{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) - \frac{\pi}{4}

None. 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 $-\tan^{-1}(1) = -\pi/4$ and $\tan^{-1}(2022)$ .

D ·

\frac{\pi}{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 $S = \sum_{k=1}^{2021} \tan^{-1}\left(\frac{1}{1 + a_k a_{k+1}}\right)$ .

02goal

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

03approach

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

04execute

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

✓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 $\tan^{-1}\left(\frac{1}{1 + a_k a_{k+1}}\right)$ 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Does the identity

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

hold without extra

\pi

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.

Answer

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

Why each option works or fails

A: $\cot^{-1}(2022) - \frac{\pi}{4}$ - 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: $\frac{\pi}{4} - \cot^{-1}(2022)$ - 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: $\tan^{-1}(2022) - \frac{\pi}{4}$ - None. 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 $-\tan^{-1}(1) = -\pi/4$ and $\tan^{-1}(2022)$ .
D: $\frac{\pi}{4} - \tan^{-1}(2022)$ - 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 $S = \sum_{k=1}^{2021} \tan^{-1}\left(\frac{1}{1 + a_k a_{k+1}}\right)$ .
goal: Evaluate the summation $\sum_{k=1}^{2021} \tan^{-1}\left(\frac{1}{1 + k(k+1)}\right)$ in terms of inverse trigonometric functions.
approach: Rewrite the numerator $1$ as $(k+1) - k$ in each term so that $\tan^{-1}\left(\frac{(k+1) - k}{1 + k(k+1)}\right) = \tan^{-1}(k+1) - \tan^{-1}(k)$ , yielding a telescoping sum.
execute: For the general term $T_k = \tan^{-1}\left(\frac{(k+1) - k}{1 + (k+1)k}\right) = \tan^{-1}(k+1) - \tan^{-1}(k)$ . Summing from $k=1$ to $2021$ gives $S = (\tan^{-1}2 - \tan^{-1}1) + (\tan^{-1}3 - \tan^{-1}2) + \dots + (\tan^{-1}2022 - \tan^{-1}2021) = \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \frac{\pi}{4}$ .
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.

Trigonometry: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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Trigonometry: JEE Main Mathematics Question with Solution

The sum telescopes via the identity tan⁡−1(x−y1+xy)=tan⁡−1x−tan⁡−1y\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}ytan−1(1+xyx−y​)=tan−1x−tan−1y, evaluating to tan⁡−1(2022)−π4\tan^{-1}(2022) - \frac{\pi}{4}tan−1(2022)−4π​.

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

Let $(a, b) \subset (0, 2\pi)$ be the largest interval for which…