Question

If $\sin 3A=1$ and $0\le A\le {90}^0$, then find the value of $\frac{1}{{\cos }^{2}A}-{\tan }^{2}A$

Answer 1

$\sin 3A=1$
$\sin 3A=\sin 90$
$3A=90$
$A={30}^{\circ }$
Now,

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

= $\frac{1}{{\cos }^{2}30˚}-{\tan }^{2}30˚$

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

= $\frac{1}{3}-\frac{4}{3}$ = $\frac{4-1}{3}$ = $1$