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
Answer & explanation
Correct answer: option 1
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,
$d b=\frac{d b}{d B} . d B \Rightarrow d b=k \cos B . d B \Rightarrow \frac{d b}{\cos B}=k d B$
∴ $\frac{d a}{\cos A}+\frac{d b}{\cos B}=k d A+k d B=k d(A+B)=k d(\pi-C)$
$\Rightarrow \quad \frac{d a}{\cos A}+\frac{d b}{\cos B}=k(0)=0$ $\left[\begin{array}{l}∵ \pi-C=\text { constant } \\ ∴ d(\pi-C)=0\end{array}\right]$