Answer: A
Given: \[ \tan\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \sqrt{3} \] Then, \[ \tan\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \tan\left(\frac{\pi}{3}\right) \] So, \[ \frac{\pi}{2} - \frac{\alpha}{2} = \frac{\pi}{3} \] Therefore, \[ -\frac{\alpha}{2} = \frac{\pi}{3} - \frac{\pi}{2} \] Or, \[ \frac{\alpha}{2} = \frac{\pi}{2} - \frac{\pi}{3} \] \[ \frac{\alpha}{2} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6} \] So, \[ \alpha = \frac{\pi}{3} \] Hence, \[ \cos\alpha = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Given: \[ \tan\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \sqrt{3} \] Then, \[ \tan\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \tan\left(\frac{\pi}{3}\right) \] So, \[ \frac{\pi}{2} - \frac{\alpha}{2} = \frac{\pi}{3} \] Therefore, \[ -\frac{\alpha}{2} = \frac{\pi}{3} - \frac{\pi}{2} \] Or, \[ \frac{\alpha}{2} = \frac{\pi}{2} - \frac{\pi}{3} \] \[ \frac{\alpha}{2} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6} \] So, \[ \alpha = \frac{\pi}{3} \] Hence, \[ \cos\alpha = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]