Respuesta :

[tex]\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -----------------------------\\\\ sec^2(x)-2=tan^2(x)\qquad [0,2\pi ) \\\\\\ \cfrac{1^2}{cos^2(x)}-2=\cfrac{sin^2(x)}{cos^2(x)}\implies \cfrac{1}{cos^2(x)}-2-\cfrac{sin^2(x)}{cos^2(x)}=0[/tex]

[tex]\bf \cfrac{1-2cos^2(x)-sin^2(x)}{cos^2(x)}=0\implies 1-2\underline{cos^2(x)}-sin^2(x) \\\\\\ 1-2\underline{[1-sin^2(x)]}-sin^2(x)=0\implies 1-2+2sin^2(x)-sin^2(x)=0 \\\\\\ -1+sin^2(x)=0\implies sin^2(x)=1\implies sin(x)=\pm\sqrt{1} \\\\\\ sin(x)=\pm 1\implies \measuredangle x=sin^{-1}(\pm1)\implies \measuredangle x= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}[/tex]
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