1621

${\displaystyle 0<x<y<\frac{\pi }{2}}$, 且 ${\displaystyle \tan y=\tan x+\frac{1}{\cos x}}$, 则 ${\displaystyle y-\frac{x}{2}=}$ _____.

根据条件式, $$\begin{array}{rl}\tan y& =\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}+\frac{1+{\tan }^2\frac{x}{2}}{1-{\tan }^2\frac{x}{2}}=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}=\tan (\frac{x}{2}+\frac{\pi }{4}),\end{array}$$ 而 ${\displaystyle y,\frac{x}{2}+\frac{\pi }{4}\in (0,\frac{\pi }{2})}$, 所以 ${\displaystyle y=\frac{x}{2}+\frac{\pi }{4}}$, 所以 ${\displaystyle y-\frac{x}{2}=\frac{\pi }{4}}$.

令 ${\displaystyle x=0}$, 则 ${\displaystyle \tan y=1}$, 所以 ${\displaystyle y=\frac{\pi }{4}}$, 所以答案是 ${\displaystyle \frac{\pi }{4}}$.