Question

If $sin3A=1sin⁡3A=1$and$0≤A≤90∘0≤A≤90∘$, find :

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

Answer 1

Given, $\sin 3A=1$
We know, $\sin 3A=\sin 90$
$\Rightarrow 3A={90}^{\circ }$
$\Rightarrow A={30}^{\circ }$
${\tan }^{2}A-\frac{1}{{\cos }^{2}A}={\tan }^{2}30-\frac{1}{{\cos }^{2}30}$

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

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