Question

If $4{\cos }^2{x}^0-1=0$ and $0\le x^0\le {90}^0$, find $\frac{1}{{\cos }^2{x}^0}-{\tan }^2{x}^0$

Answer 1

$4co{s}^2{x}^{\circ }-1=0$
$co{s}^2{x}^{\circ }=\frac{1}{4}$
$cos{x}^{\circ }=\frac{1}{2}$
$cos{x}^{\circ }=cos{60}^{\circ }$
$x={60}^{\circ }$
Now, $\frac{1}{co{s}^{2}x}-ta{n}^x$
= $se{c}^{2}x-ta{n}^{2}x$
= $se{c}^{2}60-ta{n}^{2}60$
= $2^2-(\sqrt{3}{)}^2$
= $4-3$
= $1$