Question

If A = ${60}^{\circ }$ verify $cos2A=\frac{1-{tan}^{2}A}{1+{tan}^{2}A}$

Answer 1

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

L.H.S$=\cos 2\times {60}^{\circ }$

$=\cos ({180}^{\circ }-{60}^{\circ })$

$=-\cos {60}^{\circ }$

$=\frac{-1}{2}$ ......$\left(1\right)$

R.H.S$=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

$=\frac{1-{\tan }^2{60}^{\circ }}{1+{\tan }^2{60}^{\circ }}$

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

$=\frac{1-3}{1+3}$

$=\frac{-2}{4}$

$=\frac{-1}{2}$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

L.H.S$=$R.H.S

Hence verified.