Question

If $A={30}^0$, then

$\sin 2A=2\sin A\cos A=\frac{2\tan A}{1+{\tan }^{2}A}$

If true enter 1 else 0

Answer 1

Given, $A={30}^{\circ }$
Now, $\sin 2A=2\sin A\cos A=\frac{2\tan A}{1+{\tan }^{2}A}$
L.H.S: $\sin 2A=\sin {60}^0=\frac{\sqrt{3}}{2}$
Middle Value : $2\sin A\cos A=2\sin {30}^0\cos {30}^0=2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}$
R.H.S: $\frac{2\tan A}{1+{\tan }^{2}A}$

= $\frac{2\tan {30}^0}{1+{\tan }^2{30}^0}$

= $\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+(\frac{1}{\sqrt{3}}{)}^2}$

= $\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}$

= $\frac{\sqrt{3}}{2}$
Thus, $L.H.S=R.H.S$