Question

If $A={30}^0$, then
${(\sin A-\cos A)}^2=1-\sin 2A$

If true enter 1 else 0

Answer 1

Given, $A={30}^{\circ }$
To Prove: ${(\sin A-\cos A)}^2=1-\sin 2A$
L.H.S: $(\sin A-\cos A{)}^2=(\sin {30}^0-\cos {30}^0{)}^2$
= $(\frac{1}{2}-\frac{\sqrt{3}}{2}{)}^2$
= $\frac{1+3-2\sqrt{3}}{4}$
= $\frac{2-\sqrt{3}}{2}$
R.H.S: $1-\sin 2A$
= $1-\sin {60}^0$
= $1-\frac{\sqrt{3}}{2}$
= $\frac{2-\sqrt{3}}{2}$
Hence, $L.H.S=R.H.S$