Question

In any $△ABC$, prove that $\frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}=\frac{a+b-2\sqrt{ab}}{a-b}$

(Hint: ${(a-b)}^2=a^2+b^2-2ab)$

Answer 1

In $△ABC,\frac{\sin A}{a}=\frac{\sin B}{b}$

$\Rightarrow \frac{\sin A}{\sin B}=\frac{a}{b}$

$\Rightarrow \frac{\sqrt{\sin A}}{\sqrt{\sin B}}=\sqrt{\frac{a}{b}}$

$\Rightarrow \frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$ ..... (Applying componendo and dividendo)

$\Rightarrow \frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\times \frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}$

$\Rightarrow \frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}=\frac{a+b-2\sqrt{ab}}{a-b}$