Question

If in a $△ABC$, the side $c$ and the angle $C$ remain constant, while the remaining elements are changed slightly, then the value of $\frac{da}{\cos A}+\frac{db}{\cos B}$ is

Answer 1

In a triangle, we have
$\frac{a}{\sin A}=$$\frac{b}{\sin B}=\frac{c}{\sin C}(sine-formula)$
$\Rightarrow$ $\frac{c}{\sin C}=K$ {assume}
$\therefore$ $\frac{a}{\sin A}=$$\frac{b}{\sin B}=K$
$\Rightarrow a=K\sin A,b=K\sin B$
Differentiating $a$ and $b$ we have
$da=K\cos A\cdot dA$ $\Rightarrow \frac{da}{\cos A}=KdA$
$db=K\cos B\cdot dB$ $\Rightarrow \frac{db}{\cos B}=KdB$
$\Rightarrow$ $\frac{da}{\cos A}+\frac{db}{\cos B}=KdA+KdB=kd(A+B)\cdots \left(i\right)$
As we know sum of angles of a triangle $={180}^{\circ }=\pi$
$\Rightarrow A+B+C=\pi$
$\Rightarrow A+B=\pi -C$
Substituting the values of $A+B$ in $\left(i\right)$,we get
$\frac{da}{\cos A}+\frac{db}{\cos B}=Kd(\pi -C)\cdots \left(ii\right)$
$\because C$ is a constant $\Rightarrow d(\pi -C)=0$
$\Rightarrow$ $\frac{da}{\cos A}+\frac{db}{\cos B}=K\times 0=0$