Question

In triangle ABC, prove the following:

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

Answer 1

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

$$\begin{gathered} \mathrm{LHS}=\frac{\sqrt{\sin \mathrm{A}}-\sqrt{\sin \mathrm{B}}}{\sqrt{\sin \mathrm{A}}+\sqrt{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}} \\ =\frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}+\sqrt{sinB}}\times \frac{\sqrt{\sin A}-\sqrt{\sin B}}{\sqrt{\sin A}-\sqrt{\sin B}} \\ =\frac{\sin A+\sin B-\left(2\times \sqrt{\sin A\sin B}\right)}{\sin A-\sin B} \end{gathered}$$

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

Then,
$$\begin{gathered} \mathrm{LHS}=\frac{{\displaystyle \frac{a}{k}}+{\displaystyle \frac{b}{k}}-2\times \sqrt{{\displaystyle \frac{a}{k}\frac{b}{k}}}}{{\displaystyle \frac{a}{k}}-{\displaystyle \frac{b}{k}}} \\ =\frac{{\displaystyle \frac{1}{k}}\left(a+b-2\times \sqrt{ab}\right)}{{\displaystyle \frac{1}{k}}\left(a-b\right)} \\ =\frac{a+b-2\sqrt{ab}}{a-b}=\mathrm{R}\mathrm{H}\mathrm{S} \\ \mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{d}. \end{gathered}$$