In triangle ABC, prove the following:

Consider the LHS of the equation $\frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{\sin B}}=\frac{a+b-2 \sqrt{a b}}{a-b}$.

$\mathrm{LHS}=\frac{\sqrt{\sin \mathrm{A}}-\sqrt{\sin \mathrm{B}}}{\sqrt{\sin \mathrm{A}}+\sqrt{\sin \mathrm{B}}}$

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

$=\frac{\sin A+\sin B-(2 \times \sqrt{\sin A \sin B})}{\sin A-\sin B}$

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

$\mathrm{LHS}=\frac{\frac{a}{k}+\frac{b}{k}-2 \times \sqrt{\frac{a}{k} \frac{b}{k}}}{\frac{a}{k}-\frac{b}{k}}$

$=\frac{\frac{1}{k}(a+b-2 \times \sqrt{a b})}{\frac{1}{k}(a-b)}$

$=\frac{a+b-2 \sqrt{a b}}{a-b}=\mathrm{RHS}$

Hence proved.