we can use pythagorean identieis
remember that [tex]sin^2(x)+cos^2(x)=1[/tex]
also remember that [tex]tan(x)=\frac{sin(x)}{cos{x}}[/tex] and [tex]sec(x)=\frac{1}{cos(x)}[/tex]
so if we take [tex]sin^2(x)+cos^2(x)=1[/tex] and divide both sides by [tex]cos^2(x)[/tex] we get
[tex]\frac{sin^2(x)}{cos^2(x)}+\frac{cos^2(x)}{cos^2(x)}=\frac{1}{cos^2(x)}[/tex]
[tex](\frac{sin(x)}{cos(x)})^2+1=(\frac{1}{cos(x)})^2[/tex]
[tex]tan^2(x)+1=sec^2(x)[/tex]
subtracting [tex]1+sec^2(x)[/tex] from both sides
[tex]tan^2(x)-sec^2(x)=-1[/tex]
now subsitute into original problem
[tex]\frac{tan^2(x)-sec^2(x)}{sin^2(x)+cos^2(x)}=[/tex]
[tex]\frac{-1}{1}=[/tex]
[tex]-1[/tex]
the answer is -1
