It took me a little while to think of the first one, but since I'm studying for a test, I decided to just try it until it worked, and I finally got it.

1) (tan^2x+1)(cos^2x-1) = -tan^2x

Using the Left Hand Side

(tan^2x+1)(cos^2x-1)

= **(sec^2x)(cost^2x-1)

= (1/cos^2x)(cos^2x-(cos^2x+sin^2x)

= (1/cos^2x)(cos^2x-cos^2x-sin^2x)

= (1/cos^2x)(-sin^2x)

= *(-sin^2x/cos^2x)

= -tan^2x

* where: tan^2x =sin^2x/cos^2x

** sec^2= 1 + tan^2x

derived from:

1 = sin^2x + cos^2x (divided by cos^2x)

2) 2+cos^2x-3cos^4x = sin^2x(2+3cos^2x)

Using the Right hand side

sin^2x(2+3cos^2x)

= 2sin^2x + 3sin^2cos^2x

= 2(1-cos^2x)+3(1-cos^2x)cos^2x

= 2-2cos^2x+3(cos^2-cos^4)

= 2-2cos^2x+3cos^2-3cos^4)

= 2+cos^2x-3cos^4x

Using the Identities:

1 = sin^2x + cos^2x