The angles of a convex pentagon are ${\theta }_1<{\theta }_2<{\theta }_3<{\theta }_4<{\theta }_5,$  the value of  $|\frac{\cos ({\theta }_1+{\theta }_2)}{\cos ({\theta }_3+{\theta }_4+{\theta }_5)}+\frac{\cos ({\theta }_1+{\theta }_2+{\theta }_3)}{\cos ({\theta }_4+{\theta }_5)}|$ is

The sum of all the angles of a convex n-gon is (n-2) $\pi$
Thus  ${\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4+{\theta }_5=3\pi$
 $\Rightarrow \cos ({\theta }_1+{\theta }_2)=\cos (3\pi -({\theta }_3+{\theta }_4+{\theta }_5))$
 $\Rightarrow \cos ({\theta }_1+{\theta }_2)=-\cos ({\theta }_3+{\theta }_4+{\theta }_5)$
And  $\Rightarrow \cos ({\theta }_1+{\theta }_2+{\theta }_3)=-\cos ({\theta }_4+{\theta }_5)$
 $\frac{\cos ({\theta }_1+{\theta }_2)}{\cos ({\theta }_3+{\theta }_4+{\theta }_5)}+\frac{\cos ({\theta }_1+{\theta }_2+{\theta }_3)}{\cos ({\theta }_4+{\theta }_5)}=-2$