Stuck on a fairly easy trig problem?

[theresa.rose]

For some reason it wasn’t specified whether you have to find x or try to prove the left side is equal to the right, I’m assuming one makes more sense than the other, I‘ve tried both and gotten nowhere. I’d really appreciate a little help!

 

[Steven G]

It is an identity.

What have you tried. Please post back.

 

[theresa.rose]

I’ve been trying to use the pythagorean identities 1 + tan^2x = sec^2x and 1 + cot^2x = csc^2x, however they don’t seem to line up with the signs in this problem. Do you have any other ideas?

 

[Dr.Peterson]

Is it [MATH]1-\frac{\tan^2(x)}{\cot^2(x)-1}=\tan(x)[/MATH] or [MATH]\frac{1-\tan^2(x)}{\cot^2(x)-1}=\tan(x)[/MATH]?

I find neither to be an identity. But either way, I would probably start by writing everything in terms of sin and cos.

 

[Steven G]

I suspect that the right hand side was to be tan^2(x)(?)