Tricky Trigonometry

Since the sum of the angles in a triangle is $\pi$ , thus $\frac{\angle A + \angle B - \pi}{2} = -\frac{\angle C}{2}$ . Since $\tan$ is an odd function, thus $\tan^2 \left(\frac{\angle A + \angle B - \pi}{2}\right) = \tan^2 \left(-\frac{\angle C}{2}\right) = \tan^2 \left(\frac{\angle C}{2}\right)$ . Using trigonometric identities, we have

$\begin{aligned} \tan^2 \left(\frac{\angle C}{2}\right) &= \frac{\sin^2 \left(\frac{\angle C}{2}\right)}{\cos^2 \left(\frac{\angle C}{2}\right)} \\ &= \frac{1 - \cos^2 \left(\frac{\angle C}{2}\right)}{\cos^2 \left(\frac{\angle C}{2}\right)} \\ &= \frac{1 - \frac{1^2}{3^2}}{\frac{1^2}{3^2}} \\ &= \frac{3^2-1^2}{1^2} \\ &= 8 \\ \end{aligned}$