If a triangle ABC, inscribed in a fixed circle, be slightly varied in such away as to have its vertices always on the circle, then $\frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}$ is equal

0

We know that

$\Rightarrow a=2 R \sin A, b=2 R \sin B$ and $c=2 R \sin C$

$\Rightarrow \frac{d a}{d A}=2 R \cos A, \frac{d b}{d B}=2 R \cos B$ and $\frac{d c}{d C}=2 R \cos C$

But, $d a=\frac{d a}{d A} d A, d b=\frac{d b}{d B} d B$ and $d c=\frac{d c}{d C} d C$

∴  $d a=2 R \cos A ~d A, d b=2 R \cos B ~d B$ and $d c=2 R \cos C ~d C$

$\Rightarrow \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}=2 R(d A+d B+d C)$

$\Rightarrow \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}=2 R d(A+B+C)=2 R d(\pi)$

$\Rightarrow \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}=2 R(0)=0$          [∵ A + B + C = $\pi$]