Question

Verify that :
$cos{60}^{\circ }=\frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}=\frac{1}{2}$

Answer 1

RHS = $\frac{1-ta{n}^2{30}^o}{1+ta{n}^2{30}^o}=\frac{1-(1/\sqrt{3}{)}^2}{1+(1/\sqrt{3}{)}^2}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{2/3}{4/3}=\frac{1}{2}$

LHS = $\cos {60}^o=\frac{1}{2}$

RHS = LHS

$\therefore cos{60}^o=\frac{1-ta{n}^2{30}^o}{1+ta{n}^2{30}^o}=\frac{1}{2}$

Hence proved.