We are given the equation \( \tan \left( \frac{\pi}{12} + 2x \right) = \cot 3x \).
Using the identity \( \cot \theta = \frac{1}{\tan \theta} \), we get: \[ \tan \left( \frac{\pi}{12} + 2x \right) = \frac{1}{\tan 3x} \] Multiplying both sides by \( \tan 3x \), we get: \[ \tan \left( \frac{\pi}{12} + 2x \right) \tan 3x = 1 \] Now solve for \( x \) by substituting values: \[ x = \frac{\pi}{12} \]
Let \[ f(t)=\int \left(\frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}\right)dt,\; t>1. \] If $f(e^{\pi/2})=-e^{\pi/2}$ and $f(e^{\pi/4})=\alpha e^{\pi/4}$, then $\alpha$ equals