If in a triangle the side a and the angle $\mathrm{A}$ remain constant, while other elements are changed slightly then 
 

$$\begin{gathered} \mathrm{b}=2\mathrm{R} \sin \mathrm{B} \Rightarrow \delta \mathrm{b}=2\mathrm{R} \cos \mathrm{B} \delta B \\ \mathrm{c}=2\mathrm{R} \sin \mathrm{c} \Rightarrow \delta \mathrm{c}=2\mathrm{R} \cos \mathrm{C} \delta C \\ \frac{\delta b}{\cos \mathrm{B}}+\frac{\delta c}{\cos \mathrm{C}}=2\mathrm{R}(\delta B+\delta C) \\ \mathrm{A}+\mathrm{B}+\mathrm{C}={180}^0 \Rightarrow 0+\delta B+\delta C=0 \\ \therefore \delta b \sec \mathrm{B}+\delta c \sec \mathrm{C}=0 \end{gathered}$$