Answer:
-1 (negative one)
Step-by-step explanation:
We operate as with a product of binomials, and once expanded we combine like terms:
[tex](tan(x)-sec(x))*(tan(x)+sec(x))=\\=tan^2(x)+tan(x)*sec(x)-sec(x)*tan(x)-sec^2(x)=\\=tan^2(x)-sec^2(x)[/tex]
and now we write tan and sec in terms of their basic trig expressions ([tex]tan(x)=\frac{sin(x)}{cos(x)}[/tex], and [tex]sec(x)=\frac{1}{cos(x)}[/tex]):
[tex]tan^2(x)-sec^2(x)=\frac{sin^2(x)}{cos^2(x)} -\frac{1}{cos^2(x)} =\\=\frac{sin^2(x)-1}{cos^2(x)}[/tex]
From the Pythagorean identity: [tex]sin^2(x)+cos^2(x)=1[/tex], we see that [tex]sin^2(x)-1=-cos^2(x)[/tex], so we replace this in the expression above, so we are able to cancel the factor [tex]cos^2(x)[/tex]:
[tex]\frac{sin^2(x)-1}{cos^2(x)}=\frac{-cos^2(x)}{cos^2(x)} =-1[/tex]
