If in a triangle ABC, the side c and the angle C remain constant, while the remaining elements are changed slightly, then $\frac{d a}{\cos A}+\frac{d b}{\cos B}$ is equal to

0

We are given that the side c and angle C remain constant.

$\frac{c}{\sin C}$ = k constant)

$\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=k$          $\left[∵ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\right]$

$\Rightarrow a=k \sin A \text { and } b=k \sin B$

$\Rightarrow \frac{d a}{d A}=k \cos A$ and $\frac{d b}{d B}=k \cos B$

Now,

$d a=\frac{d a}{d A} . d A \Rightarrow d a=k \cos A . d A \Rightarrow \frac{d a}{\cos A}=k d A$

and,

$d b=\frac{d b}{d B} . d B \Rightarrow d b=k \cos B . d B \Rightarrow \frac{d b}{\cos B}=k d B$

∴   $\frac{d a}{\cos A}+\frac{d b}{\cos B}=k d A+k d B=k d(A+B)=k d(\pi-C)$

$\Rightarrow \quad \frac{d a}{\cos A}+\frac{d b}{\cos B}=k(0)=0$        $\left[\begin{array}{l}∵ \pi-C=\text { constant } \\ ∴ d(\pi-C)=0\end{array}\right]$