Respuesta :
[tex]\bf \cfrac{sec(x)sin(x)+cos(\pi -x)}{1+sec(x)}\\\\
-----------------------------\\\\
sec(\theta)=\cfrac{1}{cos(\theta)}\qquad \qquad cos(\pi )=-1\qquad sin(\pi )=0
\\\\
cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})\\\\
-----------------------------\\\\
\cfrac{\frac{1}{cos(x)}sin(x)+[cos(\pi )cos(x)+sin(\pi )sin(x)]}{1+\frac{1}{cos(x)}}
\\\\\\
\cfrac{\frac{sin(x)}{cos(x)}+[-1\cdot cos(x)+0\cdot sin(x)]}{\frac{cos(x)+1}{cos(x)}}[/tex]
[tex]\bf \cfrac{\frac{sin(x)}{cos(x)}+[-cos(x)]}{\frac{cos(x)+1}{cos(x)}}\implies \cfrac{\frac{sin(x)-cos^2(x)}{cos(x)}}{\frac{cos(x)+1}{cos(x)}} \\\\\\ \cfrac{sin(x)-cos^2(x)}{cos(x)}\cdot \cfrac{cos(x)}{cos(x)+1}\implies \cfrac{sin(x)-cos^2(x)}{cos(x)+1}[/tex]
[tex]\bf \cfrac{\frac{sin(x)}{cos(x)}+[-cos(x)]}{\frac{cos(x)+1}{cos(x)}}\implies \cfrac{\frac{sin(x)-cos^2(x)}{cos(x)}}{\frac{cos(x)+1}{cos(x)}} \\\\\\ \cfrac{sin(x)-cos^2(x)}{cos(x)}\cdot \cfrac{cos(x)}{cos(x)+1}\implies \cfrac{sin(x)-cos^2(x)}{cos(x)+1}[/tex]
