How do you prove the trig identity `(sinucosu)/(cos^2u-sin^2u)=tanu/(1-tan^2u)`?

This one's quite easy, when you know what to do.

We start with `(sinucosu)/(cos^2u-sin^2u)` and we want to turn it into `tanu/(1-tan^2u)`.

For now, let's just look at the numerator: `sinucosu.` What can we do to turn this into the other numerator `(tanu)?`

Remember that `tanu=sinu/cosu`. So, the question becomes: how do we turn `sinucosu` into `sinu/cosu?`

The answer is to multiply it by `1/cos^2u`, since:

`sinucosu xx 1/color(blue)(cos^2u)=(sinu cancel cosu)/color(blue)(cosu cancel cosu)`

`color(white)(sinucosu xx 1/(cos^2u))=sinu/cosu`

If we want to multiply the numerator by `1/cos^2u`, we have to do the same thing to the denominator. That way, we're essentially multiplying the whole left side by a fancy version of `1` (that is, `(1//cos^2u)/(1//cos^2u)`), and thus we're not changing its value.

Let's see what happens when we do this:

`(sinucosu)/(cos^2u-sin^2u)=(sinucosu)/(cos^2u-sin^2u) xx color(blue)((1//cos^2u)/(1//cos^2u))`

`color(white)((sinucosu)/(cos^2u-sin^2u))=(sinu/cosu)/(" "(cos^2u-sin^2u)/cos^2u" ")`

`color(white)((sinucosu)/(cos^2u-sin^2u))=(tanu)/(cos^2u/cos^2u-sin^2u/cos^2u)`

`color(white)((sinucosu)/(cos^2u-sin^2u))=(tanu)/(1-(sinu/cosu)^2)`

`color(white)((sinucosu)/(cos^2u-sin^2u))=(tanu)/(1-tan^2u)`

And hey, look—we did it! All we had to do was focus on changing the numerator, and not get caught up in trying to change the fraction as a whole.

Note: It's not a guarantee, but when doing these trig proofs, a good rule of thumb is to look for what you can do rather than what you need to do.