Love for trigonometry

Note that $\begin{aligned} \cos2\theta & = \; 1 - 2\sin^2\theta \; = \; \tfrac14\sqrt{10 + 2\sqrt{5}} \\ \cos4\theta & = \; 2\cos^22\theta - 1 \; = \; \tfrac14(1 + \sqrt{5}) \; = \; \cos\tfrac15\pi \end{aligned}$ Assuming that $0 < \theta < \tfrac12\pi$ we have $4\theta = \tfrac15\pi\,,\,2\pi - \tfrac15\pi$ and hence $\theta = \tfrac{1}{20}\pi\,,\,\tfrac{9}{20}\pi$ . Since we know that $\cos2\theta > 0$ we deduce that $\theta = \tfrac{1}{20}\pi$ , so that $\tan\tfrac52\theta \; = \; \tan\tfrac18\pi \; = \; \sqrt{2}-1$ making the answer $\boxed{2}$ .