Question

$△ABC$ is not a right angled and is inscribed in a fixed circle. If a, A,b,B be slightly varied keeping c, C fixed then $\frac{\delta a}{cosA}+\frac{\delta b}{cosB}$ =

• A. $2$
• B. $1$
• C. $0$
• D. $5$

Answer 1

The correct option is C $0$

In $\Delta ABC$,

$a=2R\sin A$ and $b=2R\sin B$

Differentiating with respect A and B respectively.

$\frac{da}{dA}=2R\cos A$ and $\frac{db}{dB}=2R\cos B$

$da=2R\cos AdA$ and $db=2R\cos BdB$

$\frac{da}{\cos A}=2RdA$ and $\frac{db}{\cos B}=2RdB$

Now, $\frac{\delta a}{\cos A}+\frac{\delta b}{\cos B}=\frac{da}{\cos A}+\frac{db}{\cos B}$

$\Rightarrow 2RdA+2RdB$

$\Rightarrow 2Rd(A+B)$

$\Rightarrow 2Rd(\pi -c)$

$\Rightarrow 2R\times 0=0$