Respuesta :
ANSWER
EXPLANATION
We want to find the solution to the equation:
[tex]\sec ^2x-2=\tan ^2x[/tex]First, let us rewrite the trigonometric terms:
[tex]\frac{1}{\cos^2x}-2=\frac{\sin ^2x}{\cos ^2x}[/tex]Multiply both sides of the equation by cos²x:
[tex]\begin{gathered} \frac{\cos^2x}{\cos^2x}-2\cos ^2x=\sin ^2x \\ 1-2\cos ^2x=\sin ^2x \end{gathered}[/tex]We have that:
[tex]\cos ^2x+\sin ^2x=1[/tex]Substitute that for 1 in the equation:
[tex]\cos ^2x+\sin ^2x-2\cos ^2x=\sin ^2x[/tex]Simplify the equation above:
[tex]\begin{gathered} \cos ^2x-2\cos ^2x=\sin ^2x-\sin ^2x \\ \Rightarrow-\cos ^2x=0 \end{gathered}[/tex]