JEE Main 2026 — Inverse Trigonometric Functions Question with Solution

The general term of the given series is $T_p={\tan }^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right)$.

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

$T_p={\tan }^{-1}\left(\frac{2^p-2^{p-1}}{1+2^p\cdot 2^{p-1}}\right)$

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

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

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

$\sum _{p=1}^{11}{T}_p=\left({\tan }^{-1}(2^1)-{\tan }^{-1}(2^0)\right)+\left({\tan }^{-1}(2^2)-{\tan }^{-1}(2^1)\right)+\cdots +\left({\tan }^{-1}(2^{11})-{\tan }^{-1}(2^{10})\right)$

Canceling the intermediate terms, we obtain:

$\sum _{p=1}^{11}{T}_p={\tan }^{-1}(2^{11})-{\tan }^{-1}(2^0)$

Since $2^0=1$ and ${\tan }^{-1}(1)=\frac{\pi }{4}$, the sum becomes:

$\sum _{p=1}^{11}{T}_p={\tan }^{-1}(2^{11})-\frac{\pi }{4}$

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

$\alpha =\frac{\pi }{4}+{\tan }^{-1}(2^{11})-\frac{\pi }{4}$

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

Taking the tangent of both sides, we get:

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

Answer: $2048$