JEE Main 2026 — Mathematics Trigonometry

The general term of the given series is $T_p = \tan^{-1}\left(\dfrac{2^{p-1}}{1 + 2^{2p-1}}\right)$Tp​=tan−1(1+22p−12p−1​).

This can be rewritten by expressing the numerator and denominator in terms of $2^p$2p and $2^{p-1}$2p−1:

$$T_p = \tan^{-1}\left(\dfrac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}}\right)$$Tp​=tan−1(1+2p⋅2p−12p−2p−1​)

Using the inverse trigonometric identity $\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$tan−1(1+xyx−y​)=tan−1x−tan−1y, we get:

$$T_p = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$$Tp​=tan−1(2p)−tan−1(2p−1)

Summing $T_p$Tp​ from $p = 1$p=1 to $11$11 gives a telescoping series:

$$\displaystyle\sum_{p=1}^{11} T_p = \left(\tan^{-1}(2^1) - \tan^{-1}(2^0)\right) + \left(\tan^{-1}(2^2) - \tan^{-1}(2^1)\right) + \dots + \left(\tan^{-1}(2^{11}) - \tan^{-1}(2^{10})\right)$$p=1∑11​Tp​=(tan−1(21)−tan−1(20))+(tan−1(22)−tan−1(21))+⋯+(tan−1(211)−tan−1(210))

Canceling the intermediate terms, we obtain:

$$\displaystyle\sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \tan^{-1}(2^0)$$p=1∑11​Tp​=tan−1(211)−tan−1(20)

Since $2^0 = 1$20=1 and $\tan^{-1}(1) = \dfrac{\pi}{4}$tan−1(1)=4π​, the sum becomes:

$$\displaystyle\sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \dfrac{\pi}{4}$$p=1∑11​Tp​=tan−1(211)−4π​

Substituting this back into the expression for $\alpha$α:

$$\alpha = \dfrac{\pi}{4} + \tan^{-1}(2^{11}) - \dfrac{\pi}{4}$$α=4π​+tan−1(211)−4π​

$$\alpha = \tan^{-1}(2^{11})$$α=tan−1(211)

Taking the tangent of both sides, we get:

$$\tan\alpha = 2^{11} = 2048$$tanα=211=2048

Answer: $2048$2048