设 $\dfrac{1+\sin x}{\cos x}=\dfrac {22}7$ 且 $\dfrac{1+\cos x}{\sin x}=\dfrac mn$，其中 $\dfrac mn$ 为最简分数，则 $m+n=$ _______．．
答案 $44$．
解析 根据题意，有\[\begin{cases} \dfrac{1+\sin x}{\cos x}=\dfrac {22}7,\\ \dfrac{1+\cos x}{\sin x}=\dfrac mn,\end{cases}\iff \begin{cases} \tan\left(\dfrac{\pi}4-\dfrac x2\right)=\dfrac7{22},\\ \tan\dfrac x2=\dfrac nm,\end{cases}\]因此\[\dfrac{1-\dfrac nm}{1+\dfrac nm}=\dfrac 7{22}\implies \dfrac nm=\dfrac{15}{29}\implies m+n=44.\]