Inverse Trigonometric FunctionsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Telescoping arctan Sum: tan α = 2048 | JEE Main 2026

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

Question
If π4+∑p=111tan⁡−1(2p−11+22p−1)=α\dfrac{\pi}{4}+\displaystyle\sum_{p=1}^{11}\tan^{-1}\left(\dfrac{2^{p-1}}{1+2^{2p-1}}\right)=\alpha, then tan⁡α\tan\alpha is equal to
Solution
Answer: 2048 (± 0.01)
Step 1: With A=2p, B=2p−1A=2^p,\ B=2^{p-1}: A−B=2p−1(2−1)=2p−1A-B=2^{p-1}(2-1)=2^{p-1}, 1+AB=1+22p−11+AB=1+2^{2p-1}:
tan⁡−1(2p−11+22p−1)=tan⁡−1(2p)−tan⁡−1(2p−1).\tan^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right)=\tan^{-1}(2^p)-\tan^{-1}(2^{p-1}).
 Step 2: Telescoping from p=1p=1 to 1111:
∑p=111[tan⁡−1(2p)−tan⁡−1(2p−1)]=tan⁡−1(211)−tan⁡−1(1).\sum_{p=1}^{11}\big[\tan^{-1}(2^p)-\tan^{-1}(2^{p-1})\big]=\tan^{-1}(2^{11})-\tan^{-1}(1).
 Step 3: tan⁡−1(1)=π4\tan^{-1}(1)=\dfrac{\pi}{4}:
α=π4+tan⁡−1(211)−π4=tan⁡−1(211)⇒tan⁡α=211=2048.\alpha=\frac{\pi}{4}+\tan^{-1}(2^{11})-\frac{\pi}{4}=\tan^{-1}(2^{11})\Rightarrow\tan\alpha=2^{11}=2048.
 Correct answer: 2048
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