Problems

Solution: \begin{align*} && \tan \left ( \tfrac14 \pi -\tfrac12 x \right) &\equiv \frac{\tan \tfrac{\pi}{4} - \tan \tfrac12 x}{1 + \tan \tfrac{\pi}{4} \tan \tfrac12 x} \\ &&&= \frac{1-\tan \tfrac12 x}{1+\tan \tfrac12 x} \\ \\ && \sec x - \tan x &= \frac{1+t^2}{1-t^2} - \frac{2t}{1-t^2} \\ &&&= \frac{(1-t)^2}{(1-t)(1+t)} \\ &&&= \frac{1-t}{1+t} \end{align*} Therefore both sides are equal to the same thing.

1. \(\tan \tfrac18 \pi = \tan(\tfrac14 \pi - \tfrac12 \tfrac14\pi) = \sec \tfrac14 \pi - \tan \tfrac14 \pi = \sqrt{2} - 1\) \begin{align*} && \tan \tfrac{11}{24} \pi &= \tan (\tfrac13 \pi +\tfrac18 \pi) \\ &&&= \frac{\tan \tfrac13 \pi +\tan \tfrac18 \pi}{1-\tan \tfrac13 \pi \tan \tfrac18 \pi} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2}-1)} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 +\sqrt{3}-\sqrt{6}} \\ \end{align*}
2. \(\,\) \begin{align*} && (\sqrt{3}-\sqrt{6}+1)(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) &= (2\sqrt{3}+\sqrt{6}+3+3\sqrt{2}) + \\ &&&\quad+(-2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-6) + \\ &&&\quad+(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{3}+\sqrt{2}-1 \end{align*}
3. \(\,\) \begin{align*} && \tan \tfrac{1}{48} \pi &= \tan (\tfrac14\pi - \tfrac{11}{48} \pi) \\ &&&= \sec \tfrac{11}{24} \pi - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1+\tan^2 \tfrac{11}{24}\pi} - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1 + (2+\sqrt{2}+\sqrt{3}+\sqrt{6})^2} - (2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{16+10\sqrt{2} + 8\sqrt{3}+6\sqrt{6}} - 2 - \sqrt2 - \sqrt3-\sqrt6 \end{align*}