$\text{ If the sides of }\mathrm{△}ABC\text{ are changed slightly but its circum radius remains constant then }$$\frac{\delta a}{\cos A}+\frac{\delta b}{\cos B}+\frac{\delta c}{\cos C}=$

$\begin{array}{l}A+B+C={180}^{\circ }\Rightarrow \delta A+\delta B+\delta C=0\\ a=2R\sin A;b=2R\sin B;c=2R\sin C\\ \delta a=2R\cos A\delta A;\delta b=2R\cos B\delta B;\\ \delta c=2R\cos C\delta C\\ \frac{\delta a}{\cos A}+\frac{\delta b}{\cos B}+\frac{\delta c}{\cos C}=2R(\delta A+\delta B+\delta C)\\ =2R\left(0\right)=0\end{array}$