How do you simplify this in terms of sine and cosine? (All I know is that the answer is \frac {1}{cos^2(x)})

tan^2x(1+cot^2x)

First, note that \text{tan}^2x \ = \ \frac{sin^2x}{cos^2x}

Then, to give that parenthetical part of your expression a different look, start with the Pythag ID...

sin^2x \ + \ cos^2x \ = \ 1

...and divide everything in that identity by sin^2x and see where that leads you.

Give that a go and see what happens.

Well, I would have first converted everything into sin and cos, but leaving 1 as it is. (because 1 has many definitions, we get less confused by it.)

Expanding using distributive property, we get \frac{sin^2x}{cos^2x} + 1

Then, combining both into a single fraction does the trick.

Thinking more about it, there is an even faster way. tanx = \frac{1}{cotx}

So, expanding makes tan^2x + 1 which is also sec^2x! And the definition of sec is 1/cos, so , directly 1/cos^2x.

Had to spread the rep arcsine :(