Question

Verify the following equalities.
$\frac{1-{\tan }^2{60}^{\circ }}{1+{\tan }^2{60}^{\circ }}=2{\cos }^2{60}^{\circ }-1$

Answer 1

$\frac{1-{\tan }^{2}60°}{1+{\tan }^{2}60°}=\frac{1-{\left(\sqrt{3}\right)}^2}{1+{\left(\sqrt{3}\right)}^2}=\frac{1-3}{1+3}=-\frac{1}{2}$

$\Rightarrow 2{\cos }^{2}60°-1=2\times {\left(\frac{1}{2}\right)}^2-1=\frac{1}{2}-1=-\frac{1}{2}$

$\therefore \frac{1-{\tan }^{2}60°}{1+{\tan }^{2}60°}=2{\cos }^{2}60°-1$

Hence, the answer is verify.