If A = B = 60^ , verify that: (A - B) = A B - A B

Question

If $A = B = 60^\circ ,$ verify that$: \sin(A - B) = \sin A \cos B - \cos A \sin B$

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Answer

$\sin (A-B)=\sin A \cos B-\cos A \sin B$
$ \text { L.H.S. : }$
$ \sin (A-B)=\sin \left(60^{\circ}-60^{\circ}\right)=\sin 0^{\circ}=0$
$ \text { R.H.S. : }$
$ \sin A \cos B-\cos A \sin B$
$ =\sin 60^{\circ} \cos 60^{\circ}-\cos 60^{\circ} \sin 60^{\circ}$
$ =\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$
$ =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$ =0$
$ \text { L.H.S }=\text { R.H.S. }$
Therefore,
$\sin (A-B)=\sin A \cos B-\cos A \sin B \text {. }$

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