Question

How do you factor `sec^4 x-2 sec^2 x tan^2 +tan^4 x`?

Answer 1

`sec(x)^4-2sec(x)^2tan(x)^2+tan(x)^4 = 1`

Explanation:

Remember the polynomial identity
`(a - b)^2= a^2-2a b + b^2`
Let `a = sec(x)^2` and `b = tan(x)^2` so we have
`sec(x)^4-2sec(x)^2tan(x)^2+tan(2)^4 =(sec(x)^2-tan(x)^2)^2`
but `sec(x)^2-tan(x)^2 = 1/(cos(x)^2)-(sin(x)^2)/(cos(x)^2)=(1-sin(x)^2)/cos(x)^2`
Also we know `sin(x)^2+cos(x)^2=1` so putting all together
we get
`sec(x)^4-2sec(x)^2tan(x)^2+tan(x)^4 = 1`