Answer: 27

Small Hint:

Set $u=\sin t+\cos t$ and express $\sin t\cos t$ in terms of $u$

Big Hint:

Both given products become half of a perfect square in $u$

Solution:

Let $u=\sin t+\cos t.$ Since $\sin t\cos t=\frac{u^2-1}{2},$ $$\begin{aligned}(1+\sin t)&(1+\cos t)\\&=\frac{(u+1)^2}{2}=\frac54.\end{aligned}$$ The feasible sign gives $u+1=\sqrt{\frac{5}{2}}.$ Therefore $$\begin{aligned}(1-\sin t)&(1-\cos t)\\&=\frac{(1-u)^2}{2}\\&=\frac{13}{4}-\sqrt{10}.\end{aligned}$$ Thus $k+m+n=10+13+4=27.$