Respuesta :
[tex]\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
cos(\theta)=\sqrt{1-sin^2(\theta)}
\\\\\\
and\qquad tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad sec(\theta)=\cfrac{1}{cos(\theta)}\\\\
-------------------------------\\\\
[/tex]
[tex]\bf tan(x)+sec(x)=1\implies \cfrac{sin(x)}{cos(x)}+\cfrac{1}{cos(x)}=1\implies \cfrac{sin(x)+1}{cos(x)}=1 \\\\\\ sin(x)+1=cos(x)\implies sin(x)+1=\boxed{\sqrt{1-sin^2(x)}} \\\\\\\ [sin(x)+1]^2=[\sqrt{1-sin^2(x)}]^2 \\\\\\ sin^2(x)+2sin(x)+1^2=1-sin^2(x) \\\\\\ 2sin^2(x)+2sin(x)=0\implies 2sin(x)[sin(x)+1]=0 \\\\\\ \begin{cases} 2sin(x)=0\\ \qquad sin(x)=0\\ \qquad \measuredangle x=0~,~\pi \\ sin(x)+1=0\\ \qquad sin(x)=-1\\ \qquad \measuredangle x=\frac{3\pi }{2} \end{cases}[/tex]
[tex]\bf tan(x)+sec(x)=1\implies \cfrac{sin(x)}{cos(x)}+\cfrac{1}{cos(x)}=1\implies \cfrac{sin(x)+1}{cos(x)}=1 \\\\\\ sin(x)+1=cos(x)\implies sin(x)+1=\boxed{\sqrt{1-sin^2(x)}} \\\\\\\ [sin(x)+1]^2=[\sqrt{1-sin^2(x)}]^2 \\\\\\ sin^2(x)+2sin(x)+1^2=1-sin^2(x) \\\\\\ 2sin^2(x)+2sin(x)=0\implies 2sin(x)[sin(x)+1]=0 \\\\\\ \begin{cases} 2sin(x)=0\\ \qquad sin(x)=0\\ \qquad \measuredangle x=0~,~\pi \\ sin(x)+1=0\\ \qquad sin(x)=-1\\ \qquad \measuredangle x=\frac{3\pi }{2} \end{cases}[/tex]
