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

Answer

$\frac{\sqrt{\sin\text{A}}-\sqrt{\sin\text{B}}}{\sqrt{\sin\text{A}}+\sqrt{\sin\text{B}}}=\frac{\text{a + b}-2\sqrt{\text{ab}}}{\text{a}-\text{b}}$
$\text{RHS}=\frac{\text{a + b}-2\sqrt{\text{ab}}}{\text{a}-\text{b}}$
$=\frac{(\sqrt{\text{a}})^2+(\sqrt{\text{b}})^2-2\sqrt{\text{ab}}}{(\sqrt{\text{a}})^2-(\sqrt{\text{b}})^2}$
$=\frac{(\sqrt{\text{a}}-\sqrt{\text{b}})^2}{(\sqrt{a})^2-(\sqrt{\text{b}})^2}$
$=\frac{(\sqrt{\text{a}}-\sqrt{\text{b}})}{(\sqrt{\text{a}}+\sqrt{\text{b}})}$
$=\frac{\big(\sqrt{\text{k}\sin\text{A}}-\sqrt{\text{k}\sin\text{B}}\big)}{\big(\sqrt{\text{k}\sin\text{A}}+\sqrt{\text{k}\sin\text{B}}\big)}$
$=\frac{\big(\sqrt{\sin\text{A}}-\sqrt{\sin\text{B}}\big)}{\big(\sqrt{\sin\text{A}}+\sqrt{\sin\text{B}}\big)}=\text{LHS}$ [taking k common and cancelling them]
Hence proved

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