If in a triangle ABC, the side a and the angle A remain constant, while the remaining elements are changed slightly, show that $\dfrac{db}{\cos B} + \dfrac{dc}{\cos C} = 0$
According to sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k \left(\text{ constant}\right)$
$\therefore$ $a = k \sin A$, $b = k \sin B$, $c = k \sin C$
$\therefore$ $\dfrac{da}{dA}= k \cos A$, $\dfrac{db}{dB} = k \cos B$, $\dfrac{dc}{dC} = k \cos C$
But $\dfrac{da}{dA} = 0$ since a is a constant (given)
Now $db = \left(\dfrac{db}{dB}\right) \times \Delta B = k \cos B \times \Delta B$
$\implies$ $\dfrac{db}{\cos B} = k \Delta B$
and $dc = \left(\dfrac{dc}{dC}\right) \times \Delta C = k \cos C \times \Delta C$
$\implies$ $\dfrac{dc}{\cos C} = k \Delta C$
$\begin{aligned}
\therefore \dfrac{db}{\cos B} + \dfrac{dc}{\cos C} & = k \Delta B + k \Delta C \\
& = k \left(\Delta B + \Delta C\right) \\
& = k \Delta \left(B+C\right) \\
& = k \Delta \left(\pi - A\right) \;\; \left[\text{Note: In a triangle, }A+B+C=\pi\right] \\
& = k \times 0 \;\; \left[\text{Note: A is a constant (given)}\right] \\
& = 0
\end{aligned}$
$\therefore$ $\dfrac{db}{\cos B}+ \dfrac{dc}{\cos C} = 0$