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　$\displaystyle0<u<\frac{\pi}{4}$，$\displaystyle0<\theta<\frac{\pi}{2}$とする．
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\item[(1)]$\displaystyle\frac{1}{\tan u}-\frac{1}{\tan2u}$を簡単にせよ．
\item[(2)]無限級数$\displaystyle\sum_{n=1}^\infty\frac{1}{2^n}\tan\frac{\theta}{2^n}$の和を求めよ．
ただし，$\displaystyle\lim_{x\to0}\frac{\tan x}{x}=1$を用いてよい．
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