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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