When trying to prove trig identities, it is often helpful to convert TAN functions into SIN/COS functions:
Proof Step 1: Start with the original equation to prove:
tan^{2}x  sin^{2}x = (tan^{2}x)(sin^{2}x)
Proof Step 2: Replace tan with sin/cos
(sin^{2}x/cos^{2}x)  sin^{2}x = (sin^{2}x/cos^{2}x)(sin^{2}x)
Proof Step 3: Obtain a common denominator on left, simplify right
(sin^{2}x  sin^{2}x cos^{2}x) / cos^{2}x = sin^{4}x / cos^{2}x
Proof Step 4: Cancel cos^{2}x from both denominators
sin^{2}x  sin^{2}x cos^{2}x = sin^{4}x
Proof Step 5: Factor out sin^{2}x from left
(sin^{2}x)(1  cos^{2}x) = sin^{4}x
Proof Step 6: Use trig identity 1cos^{2}x = sin^{2}x
(sin^{2}x)(sin^{2}x) = sin^{4}x
Proof Step 7: Simplify, DONE.
sin^{4}x = sin^{4}x
11/18/2012

Michael B.
Comments
Could you remind me how you factored out sin^2x from the left in proof step 5? Other than that, everything else was very helpful, it makes so much more sense.
look at (sin^2(x)sin^2(x)cos^2(x)) as (22x). Both terms have a two therefore can be factored out, 2(1x).
Angel's answer is generally correct, if a bit confusing. I would have said it this way: think of (sin^{2}x  sin^{2}x cos^{2}x) as (a  ab) where a=sin^{2}x and b=cos^{2}x. Now, when 'factoring out' we are looking for an expression that is common in both terms 'a' and 'ab'. The common term is simply 'a', so we can factor it out. Factoring it out is the opposite of the 'distribution' property  that is, it is distribution in reverse. So for each term, we can ask "a times what is equal to the term?" So for the first term, a, we say "a times what is equal to a?" and the answer is 1. For the second term, "a times what is equal to ab?" and the answer is b. So: (a  ab) = a(1  b).
Now since we said a=sin2x and b=cos^{2}x, we know that:
sin^{2}x  sin^{2}x cos^{2}x = sin^{2}x (1  cos^{2}x)