Question
If $A=30^{\circ}$ and $B=60^{\circ}$, verify that: $\frac{\sin (A-B)}{\sin A \cdot \sin B}=\cot B-\cot A$
✓
Answer
$A = 30^\circ$ and $B = 60^\circ$
$\text{L.H.S.}$
$=\frac{\sin (A-B)}{\sin A \cdot \sin B}$
$=\frac{\sin \left(30^{\circ}-60^{\circ}\right)}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\sin 30^{\circ}}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\frac{1}{2}}{\frac{1}{2} \times \frac{\sqrt{3}}{2}}$
$=-\frac{2}{\sqrt{3}}$
$\text{R.H.S.}$
$= \cot B - \cot A$
$= \cot 60^\circ -\cot 30^\circ$
$=\frac{1}{\sqrt{3}}-\sqrt{3}$
$=\frac{1-3}{\sqrt{3}}$
$=-\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{\sin (A-B)}{\sin A \cdot \sin B}=\cot B-\cot A$
$\text{L.H.S.}$
$=\frac{\sin (A-B)}{\sin A \cdot \sin B}$
$=\frac{\sin \left(30^{\circ}-60^{\circ}\right)}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\sin 30^{\circ}}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\frac{1}{2}}{\frac{1}{2} \times \frac{\sqrt{3}}{2}}$
$=-\frac{2}{\sqrt{3}}$
$\text{R.H.S.}$
$= \cot B - \cot A$
$= \cot 60^\circ -\cot 30^\circ$
$=\frac{1}{\sqrt{3}}-\sqrt{3}$
$=\frac{1-3}{\sqrt{3}}$
$=-\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{\sin (A-B)}{\sin A \cdot \sin B}=\cot B-\cot A$
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