If $I_n=\int\limits_0^{\pi/4}\tan^nx\, dx$ then $I_{2024} +I_{2026}$ is equal to:

$\frac{1}{2025}$

The correct answer is Option (1) → $\frac{1}{2025}$

$I_n=\displaystyle \int_{0}^{\pi/4} \tan^n x \, dx$

The known reduction identity is:

$I_n + I_{n+2} = \frac{1}{n+1}$

Substitute $n=2024$:

$I_{2024} + I_{2026} = \frac{1}{2025}$

$\frac{1}{2025}$