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orkdork · Oct 30, 2009, 12:36 PM

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

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

ArcSine · Oct 30, 2009, 03:21 PM

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.

Unknown008 · Oct 31, 2009, 12:31 AM

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 :(