Recall the important Pythagorean Identity 1+tan^2x=sec^2x1+tan2x=sec2x. We will be needing it for this problem.
Let's start with the numerator:
sec^4x-1sec4x−1
Note that this can be rewritten as:
(sec^2x)^2-(1)^2(sec2x)2−(1)2
This fits the form of a difference of squares, a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b), with a=sec^2xa=sec2x and b=1b=1. It factors into:
(sec^2x-1)(sec^2x+1)(sec2x−1)(sec2x+1)
From the identity 1+tan^2x=sec^2x1+tan2x=sec2x, we can see that subtracting 11 from both sides gives us tan^2x=sec^2x-1tan2x=sec2x−1. We can therefore replace sec^2x-1sec2x−1 with tan^2xtan2x:
(sec^2x-1)(sec^2x+1)(sec2x−1)(sec2x+1)
->(tan^2x)(sec^2x+1)→(tan2x)(sec2x+1)
Let's check out the denominator:
sec^4x+sec^2xsec4x+sec2x
We can factor out a sec^2xsec2x:
sec^4x+sec^2xsec4x+sec2x
->sec^2x(sec^2x+1)→sec2x(sec2x+1)
There isn't much more we can do here, so let's look at what we have now:
((tan^2x)(sec^2x+1))/((sec^2x)(sec^2x+1))(tan2x)(sec2x+1)(sec2x)(sec2x+1)
We can do some canceling:
((tan^2x)cancel((sec^2x+1)))/((sec^2x)cancel((sec^2x+1))
->tan^2x/sec^2x
Now we rewrite this using only sines and cosines and simplify:
tan^2x/sec^2x
->(sin^2x/cos^2x)/(1/cos^2x)
->sin^2x/cos^2x*cos^2x
->sin^2x/cancel(cos^2x)*cancel(cos^2x)=sin^2x
