恒等变换

$\mathrm{设}0<x<y<\frac{\pi }{2},\mathrm{且}\tan y=\tan x+\frac{1}{\cos x},\mathrm{则}y-\frac{x}{2}=()$

$\mathrm{常}\mathrm{规}\mathrm{思}\mathrm{路}:\mathrm{切}\mathrm{化}\mathrm{弦}，\mathrm{统}\mathrm{一}\mathrm{三}\mathrm{角}\mathrm{函}\mathrm{数}\mathrm{名}$
$\tan y=\tan x+\frac{1}{\cos x}\mathrm{可}\mathrm{化}\mathrm{为}$
$\frac{\sin y}{\cos y}=\frac{\sin x+1}{\cos x}$
$\mathrm{整}\mathrm{理}\mathrm{可}\mathrm{得}\sin y\cos x=\cos y\sin x+\cos y$
$\mathrm{即}\sin y\cos x-\cos y\sin x=\cos y\Rightarrow$
$\sin (y-x)=\cos x=\sin (\frac{\pi }{2}-x)$

$\because 0<x<y<\frac{\pi }{2}$
$\therefore 0<y-x<\frac{\pi }{2}$
$0<\frac{\pi }{2}-x<\frac{\pi }{2}$
$\therefore y-x=\frac{\pi }{2}-x$
$y-2x=\frac{\pi }{2}$
$\mathrm{结}\mathrm{果}\mathrm{为}\frac{\pi }{4}$

$\mathrm{利}\mathrm{用}\mathrm{半}\mathrm{角}\mathrm{公}\mathrm{式}\tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=\frac{1-\cos \theta }{\sin \theta }$
$\tan y=\frac{1+\sin x}{\cos x}=\frac{1-\cos (x+\frac{\pi }{2})}{\sin (x+\frac{\pi }{2})}=\tan (\frac{x}{2}+\frac{\pi }{4})$
$\because 0<x<y<\frac{\pi }{2}$
$\therefore 0<\frac{x}{2}+\frac{\pi }{4}<\frac{\pi }{2}$
$\therefore y=\frac{x}{2}+\frac{\pi }{4}$