Question

If $A=B={45}^0$, then

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

If true enter 1 else 0

Answer 1

Given, $A=B={45}^{\circ }$
Now, $\sin (A-B)=\sin A\cos B-\cos A\sin B$
L.H.S: $\sin (A-B)=\sin ({45}^0-{45}^0)=\sin 0^0=0$
R.H.S : $\sin A\cos B-\cos A\sin B$
= $\sin {45}^0\cos {45}^0-\cos {45}^0\sin {45}^0$
= $\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$
= $\frac{1}{2}-\frac{1}{2}$
= $0$