Answer:
-1 (negative one)
Explanation:
We operate as with a product of binomials, and once expanded we combine like terms:
![(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)](https://img.qammunity.org/2020/formulas/mathematics/high-school/lrd83qar7n8mlz3n1g8a1o5grcrgusnnj5.png)
and now we write tan and sec in terms of their basic trig expressions (
, and
):
![tan^2(x)-sec^2(x)=(sin^2(x))/(cos^2(x)) -(1)/(cos^2(x)) =\\=(sin^2(x)-1)/(cos^2(x))](https://img.qammunity.org/2020/formulas/mathematics/high-school/426hkthv7nehrlgio0ac225vd08xg8pmbq.png)
From the Pythagorean identity:
, we see that
, so we replace this in the expression above, so we are able to cancel the factor
:
![(sin^2(x)-1)/(cos^2(x))=(-cos^2(x))/(cos^2(x)) =-1](https://img.qammunity.org/2020/formulas/mathematics/high-school/jn9a85jtdud8t0x1n8crat0t6ix62s77w8.png)