由 $\tan \left(\dfrac {\alpha}{2}+\dfrac {\pi}{4}\right)=3+2\sqrt 2$ 可得$$\tan{\dfrac{\alpha}{2}}=\dfrac{\sqrt 2}{2}.$$所以$$\begin{split}\dfrac {1-\sin \alpha}{\cos \alpha}&=\dfrac{\sin^2{\dfrac{\alpha}{2}}+\cos^2{\dfrac{\alpha}{2}}-2\sin{\dfrac{\alpha}{2}}\cos{\dfrac{\alpha}{2}}}{\cos^2{\dfrac{\alpha}{2}}-\sin^2{\dfrac{\alpha}{2}}}\\&=\dfrac{\tan^2{\dfrac {\alpha}{2}}+1-2\tan{\dfrac{\alpha}{2}}}{1-\tan^2{\dfrac{\alpha}{2}}}\\&=3-2\sqrt2.\end{split}$$