Question

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

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

If true enter 1 else 0

Answer 1

Given, $A={30}^{\circ }$
Now, $\cos 2A={\cos }^{2}A-{\sin }^{2}A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$
L.H.S : $\cos 2A=\cos 2\times 30=\cos 60=\frac{1}{2}$
Middle Side: ${\cos }^{2}A-si{n}^{2}A={\cos }^{2}30-si{n}^{2}30=(\frac{\sqrt{3}}{2}{)}^2-(\frac{1}{2}{)}^2$
= $\frac{3}{4}-\frac{1}{4}$
= $\frac{2}{4}$
= $\frac{1}{2}$
R.H.S: $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$
= $\frac{1-{\tan }^{2}30}{1+{\tan }^{2}30}$
= $\frac{1-(\frac{1}{\sqrt{3}}{)}^2}{1+(\frac{1}{\sqrt{3}}{)}^2}$
= $\frac{\frac{2}{3}}{\frac{4}{3}}$
= $\frac{1}{2}$
Hence, $L.H.S=M.H.S=R.H.S$