Thread: Prove this identity (quick!)

Prove this identity:

[ (cos x) / (1 - tan x) ] + [ (sin x) / (1 - cot x) ] = cos x + sin x

So far I was able to put everything in terms of sin x and cos x
I understand that you have to multiply each dominator and numerator to get the common dominator.
Anyway, I got to this portion of it:

[ (cos^2 x) / (cos - sin) ] + [ (sin^2 x) / (sin - cos) ] = cos x + sin x

I do have a solution I could copy, but I'd rather understand how to do the problem. It's quite frustrating.

Originally Posted by Skynt

Prove this identity:

[ (cos x) / (1 - tan x) ] + [ (sin x) / (1 - cot x) ] = cos x + sin x

So far I was able to put everything in terms of sin x and cos x
I understand that you have to multiply each dominator and numerator to get the common dominator.
Anyway, I got to this portion of it:

[ (cos^2 x) / (cos - sin) ] + [ (sin^2 x) / (sin - cos) ] = cos x + sin x

I do have a solution I could copy, but I'd rather understand how to do the problem. It's quite frustrating.

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Skynt,
You are almost there.
work little more on dominator then it will become
( sin^2x - cos^2x ) / ( sin x - cos x )

and then
[ (sin x + cos x) ( sin x - cos x) ] / ( sin x - cos x)

and you start to cancel ( sin x - cos x ) at numinator and dominator. the only thing left is ( sin x + cos x )