Question

$\frac{1}{co{s}^2{x}^{\circ }}-ta{n}^2{x}^{\circ }$

Answer 1

$4{\cos }^2{x}^0-1=0$
${\cos }^{2}x=\frac{1}{4}$
$\cos x=\pm \frac{1}{2}$
Neglect the negative value,
$\cos x=\frac{1}{2}$
$\cos x=\cos 60$
$x={60}^{\circ }$
Now, $\frac{1}{{\cos }^{2}x}-{\tan }^{2}x={\sec }^{2}x-{\tan }^{2}x$
= ${\sec }^{2}60-{\tan }^{2}60$
= $2^2-(\sqrt{3}{)}^2$
= $4-3$
= $1$