Question

If in a triangle $ABC$, the side $c$ and the length $C$ remain constant, while the remaining elements are changed slightly, show that $\frac{da}{\cos A}+\frac{db}{\cos B}=0$

Answer 1

In a triangle, we have

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\left[\text{By sine-formula}\right]$

Given that side c and angle C is constant.

$\therefore \frac{c}{\sin C}=K\left(\text{assume}\right)$

$\therefore \frac{a}{\sin A}=\frac{b}{\sin B}=K$

$\Rightarrow a=K\sin A,b=K\sin B$

Differentiating a and b, we have

$da=K\cos AdA\Rightarrow \frac{da}{\cos A}=KdA$

$db=K\cos BdB\Rightarrow \frac{db}{\cos B}=KdB$

$\therefore \frac{da}{\cos A}+\frac{db}{\cos B}=KdA+KdB=Kd(A+B).....\left(1\right)$

As we know that, sum of all the angles of a triangle is 180°.

$\therefore A+B+C=180°=\pi$

$\Rightarrow A+B=\pi -C$

Substituting the value of $(A+B)$ in ${eq}^n\left(1\right)$, we get

$\frac{da}{\cos A}+\frac{db}{\cos B}=Kd(\pi -C).....\left(2\right)$

$\because C=\text{consatnt}\left[\text{Given}\right]$

$\therefore (\pi -C)=\text{constant}$

$\Rightarrow d(\pi -C)=0$

Substituting the value of $d(\pi -C)$ in ${eq}^n\left(\right)$, we get

$\frac{da}{\cos A}+\frac{db}{\cos B}=K\times 0=0$

Hence proved that $\frac{da}{\cos A}+\frac{db}{\cos B}=0$.