Question

If $A={30}^{\circ }$, verify that $\cos 2A=\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}$.

Answer 1

$A=30\Rightarrow 2A=60$
$\cos 2A=\cos {60}^{\circ }=\frac{1}{2}$
Also, $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}=\frac{1-{\tan }^2{30}^{\circ }}{1+{\tan }^2{30}^{\circ }}=\frac{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}$
$=\frac{(1-\frac{1}{3})}{(1+\frac{1}{3})}=\frac{\left(\frac{2}{3}\right)}{\left(\frac{4}{3}\right)}=(\frac{2}{3}\times \frac{3}{4})=\frac{1}{2}$
Hence, $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$.