Question

If $A={60}^0$ and $B={30}^0$;

$\frac{\sin (A-B)}{\sin A\sin B}=\cot B-\cot A$

If true enter 1 else 0

Answer 1

Given: $A={60}^{\circ },B={30}^{\circ }$
Now, $\frac{\sin (A-B)}{\sin A\sin B}=\cot B-\cot A$
L.H.S: $\frac{\sin (A-B)}{\sin A\sin B}$
= $\frac{\sin (60-30)}{\sin 60\sin 30}$
= $\frac{\sin 30}{\sin 60\sin 30}$
= $\frac{1}{\sin 60}$
= $\frac{1}{\frac{\sqrt{3}}{2}}$
= $\frac{2}{\sqrt{3}}$
R.H.S : $\cot B-\cot A$
= $\cot 30-\cot 60$
= $\sqrt{3}-\frac{1}{\sqrt{3}}$
= $\frac{3-1}{\sqrt{3}}$
= $\frac{2}{\sqrt{3}}$
Thus, $L.H.S=R.H.S$