Mathematics - Quadratic Equation Question with Solution | TestHub

$\because a(b-c)+b(c-a)+c(a-b)=0$ $\therefore \mathrm{x}=1$ is a root of the equation

$$a(b-c)x^2+b(c-a)x+c(a-b)=0$$

Then, other root $=1(\because$ roots are equal) $\therefore \alpha \times \beta =\frac{c(a-b)}{a(b-c)}$ $\Rightarrow ab-ac=ca-bc$ $\therefore b=\frac{2ac}{a+c}$

$\therefore a,b,c$ are in HP Then, $\frac{1}{\mathrm{a}},\frac{1}{\mathrm{b}},\frac{1}{\mathrm{c}}$ are in AP.

$\Rightarrow \frac{\mathrm{s}}{\mathrm{a}},\frac{\mathrm{s}}{\mathrm{b}},\frac{\mathrm{s}}{\mathrm{c}}$ are in AP

$\Rightarrow \frac{\mathrm{s}}{\mathrm{a}}-1,\frac{\mathrm{s}}{\mathrm{b}}-1,\frac{\mathrm{s}}{\mathrm{c}}-1$ are in AP.

$\Rightarrow \frac{(\mathrm{s}-\mathrm{a})}{\mathrm{a}},\frac{(\mathrm{s}-\mathrm{b})}{\mathrm{b}},\frac{(\mathrm{s}-\mathrm{c})}{\mathrm{c}}$ are in AP.

Multiplying in each by $\frac{abc}{(s-a)(s-b)(s-c)}$ Then $\frac{\mathrm{bc}}{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})},\frac{\mathrm{ca}}{(\mathrm{s}-\mathrm{c})(\mathrm{s}-\mathrm{a})},\frac{\mathrm{ab}}{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}$ are in AP. $\Rightarrow \frac{(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}{\mathrm{bc}},\frac{(\mathrm{s}-\mathrm{c})(\mathrm{s}-\mathrm{a})}{\mathrm{ca}},\frac{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}{\mathrm{ab}}$ are in HP. Or ${\sin }^2\left(\frac{\mathrm{A}}{2}\right),{\sin }^2\left(\frac{\mathrm{B}}{2}\right),{\sin }^2\left(\frac{\mathrm{C}}{2}\right)$ are in HP