Trigonometry JEE 2026: Arctan Sum and Endpoint Checks

What is the answer to the Trigonometry JEE 2026 arctan sum?

The answer is 2048, not 1024: the surviving positive endpoint comes from the eleventh power of two. This trigonometry JEE 2026 problem is identified in the supplied bank as a JEE Main inverse-trigonometric-functions question, tagged hard on the bank’s own scale.

Eleven inverse-tangent angles, indexed from 1 through 11, define the angle below. Find its tangent.

α = \frac{π}{4} + \sum_{p = 1}^{11} \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) .

Here, inverse tangent does not mean the reciprocal of tangent. The two notations below mean the same thing: $\tan^{- 1} x = \arctan x .$

The source classifies the question as Short Answer, but also supplies these comparison options. They are not evidence of the exam’s question format.

The numerical answer corresponds to supplied option B:

\tan α = 2048 .

How do you turn each summand into an arctan difference?

Reconstruct two consecutive powers of two: their difference gives the numerator, and their product gives the denominator’s power. This is the official solution’s recognition step. The identity is branch-safe here because the difference between the two inverse-tangent angles lies inside the principal range.

The target structure and choices are:

\frac{u - v}{1 + u v}, u = 2^{p}, v = 2^{p - 1} .

Check both parts rather than guessing from the numerator:

u - v = 2^{p} - 2^{p - 1} = 2^{p - 1},

u v = 2^{p} \times 2^{p - 1} = 2^{2 p - 1} .

Name the two angles and apply the tangent-difference formula:

A = \arctan (2^{p}), B = \arctan (2^{p - 1}) .

\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{2^{p - 1}}{1 + 2^{2 p - 1}} .

Both arguments are positive, and the first is larger. Since principal arctan is increasing,

0 < B < A < \frac{π}{2} ⟹ 0 < A - B < \frac{π}{2} .

The difference already lies inside the principal arctan interval:

(- \frac{π}{2}, \frac{π}{2}) .

Taking principal arctan therefore introduces no multiple of pi:

\arctan (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = \arctan (2^{p}) - \arctan (2^{p - 1}) .

Do not treat this arctan-difference identity as unconditional for arbitrary real arguments. The angle-range check is part of the solution.

Which endpoints survive when the eleven terms cancel?

Only the positive angle in the eleventh bracket and the negative angle in the first bracket survive. Every intermediate angle occurs twice, once with each sign. Write the boundary brackets rather than mentally counting powers.

Denote the sum without the initial angle by:

S = \sum_{p = 1}^{11} [\arctan (2^{p}) - \arctan (2^{p - 1})] .

Expand the first two and final two brackets:

S = & [\arctan 2 - \arctan 1] & + [\arctan 4 - \arctan 2] & + \dots & + [\arctan 1024 - \arctan 512] & + [\arctan 2048 - \arctan 1024] .

Each positive intermediate angle cancels its negative occurrence in the following bracket. The first negative angle has no earlier partner; the last positive angle has no later partner.

Endpoint audit: read the positive endpoint from the upper index and the negative endpoint from the lower index before simplifying either exponent.

p = 11 : + \arctan (2^{11}),

p = 1 : - \arctan (2^{1 - 1}) .

The surviving endpoints give:

S = \arctan (2^{11}) - \arctan (2^{0}) = \arctan 2048 - \arctan 1 .

How does the initial angle give the final tangent?

The initial angle cancels the lower endpoint exactly. No tangent-addition calculation is needed after telescoping.

\arctan 1 = \frac{π}{4} .

Substitute this value, then apply tangent:

α = \frac{π}{4} + \arctan 2048 - \arctan 1 = \arctan 2048 .

\tan α = \tan (\arctan 2048) = 2048 .

The numerical answer is 2048; B is only the corresponding supplied bank option. Since the remaining argument exceeds one, the angle satisfies:

\frac{π}{4} < α < \frac{π}{2} .

The bank’s 120-second expected time is a practice benchmark based on its listed estimate. It is not a measured student average or a guaranteed solving time.

How can an endpoint error produce 1024?

Using the numerator’s exponent as the positive endpoint’s exponent produces 1024. The numerator comes from subtracting two arguments; it is not itself the larger reconstructed argument.

The incorrect endpoint calculation is:

S_{wrong} = \arctan (2^{10}) - \arctan 1 .

This would give supplied option A:

α_{wrong} = \frac{π}{4} + S_{wrong} = \arctan 1024,

$\tan α_{wrong} = 1024 .$ Keep the roles separate:

Numerator = 2^{p - 1}, positive arctan argument = 2^{p} .

At the final index, the correct difference is:

p = 11 : \arctan 2048 - \arctan 1024 .

The mistaken positive endpoint is: $\arctan 1024 .$

That angle is actually subtracted in the final bracket. It cancels the positive angle from the preceding bracket.

Repair the calculation by writing both boundary differences first:

p = 1 : \arctan 2 - \arctan 1,

p = 11 : \arctan 2048 - \arctan 1024 .

Only then cancel the middle terms. This prevents the numerator’s exponent from replacing the positive endpoint’s exponent.

Which three practice questions test the same method?

Use this sequence to test endpoint handling, recognition with another base, and finally a principal-value check where a pi correction is necessary. All three are original practice questions, not additional verified PYQs. Attempt each before reading its worked check.

Question 1: Find the tangent of the angle defined below.

β = \frac{π}{4} + \sum_{p = 1}^{6} \arctan (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) .

The same consecutive-power differences apply. The first negative argument is one; the sixth positive argument is 64: $S = \arctan 64 - \arctan 1 .$

β = \arctan 64, \tan β = 64 .

Question 2: Find the tangent when the base changes to three.

γ = \frac{π}{4} + \sum_{p = 1}^{4} \arctan (\frac{2 \times 3^{p - 1}}{1 + 3^{2 p - 1}}) .

The factor of two in the numerator comes from the subtraction. The product supplies the denominator’s power:

3^{p} - 3^{p - 1} = 2 \times 3^{p - 1}, 3^{p} \times 3^{p - 1} = 3^{2 p - 1} .

The positive, ordered arguments give the same branch-safe difference:

S & = \sum_{p = 1}^{4} [\arctan (3^{p}) - \arctan (3^{p - 1})] & = \arctan 81 - \arctan 1 .

γ = \arctan 81, \tan γ = 81 .

Question 3: Evaluate this sum of angles, not merely its tangent. $θ = \arctan 2 + \arctan 3 .$

The tangent-addition formula gives:

\tan θ = \frac{2 + 3}{1 - 6} = - 1 .

Both component angles satisfy:

\frac{π}{4} < \arctan 2 < \frac{π}{2}, \frac{π}{4} < \arctan 3 < \frac{π}{2} .

Their sum therefore lies in the second quadrant:

\frac{π}{2} < θ < π .

That interval fixes the required angle. Taking principal arctan alone would give the wrong value:

θ = \frac{3 π}{4}, θ \neq - \frac{π}{4} .

Here, a pi correction is necessary: $θ = \arctan (- 1) + π .$

On your next attempt, write the endpoint pair for the first two questions and the angle interval for the third before checking the answers.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to the Trigonometry JEE 2026 arctan-sum problem?

The required tangent is 2048. The eleven-term sum telescopes to arctan(2048) − arctan(1), and the initial π/4 cancels arctan(1). Thus α = arctan(2048), giving tan α = 2048.

How do you convert each term into an arctan difference?

Choose u = 2^p and v = 2^(p−1), so u − v = 2^(p−1) and uv = 2^(2p−1). The summand becomes arctan(2^p) − arctan(2^(p−1)). This is branch-safe because the difference lies between 0 and π/2.

Why is the answer 2048 and not 1024?

The final positive arctan argument is 2^11 = 2048, not the numerator's power 2^10 = 1024. The last bracket is arctan(2048) − arctan(1024), so arctan(1024) cancels with the preceding bracket. Writing the first and last brackets prevents this endpoint error.

When is a pi correction needed in an arctan calculation?

A correction is needed when the angle being recovered lies outside the principal arctan interval (−π/2, π/2). For example, arctan(2) + arctan(3) lies in the second quadrant and has tangent −1. Its value is therefore arctan(−1) + π = 3π/4, not −π/4.

Trigonometry JEE 2026: Arctan Sum by Telescoping