If a triangle $A B C,$ 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

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none of these

answer is A.

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Detailed Solution

We know that

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

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

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

$\begin{array}{ll} ∴ d a = 2 R \cos A d A, d b = 2 R \cos B d B and d c = 2 R \cos C d C \\ \Rightarrow \frac{d a}{\cos A} + \frac{d b}{\cos B} + \frac{d c}{\cos C} = 2 R (d A + d B + d C) \\ \Rightarrow \frac{d a}{\cos A} + \frac{d b}{\cos B} + \frac{d c}{\cos C} = 2 R d (A + B + C) = 2 R d (π) \\ \Rightarrow \frac{d a}{\cos A} + \frac{d b}{\cos B} + \frac{d c}{\cos C} = 2 R (0) = 0 [∵ A + B + C = π] \end{array}$