# Thread: Help with trigonometric identities.

1. ## Help with trigonometric identities.

I need some help proving 1/(cscx - cotx) = (1 + cosx)/sinx. I just started dealing with these today and I don't fully comprehend how I am supposed to transform 1/(cscx - cotx) into (1 + cosx)/sinx, so a very detailed explaination would be ideal. Thank you in advance.

2. $\displaystyle \frac{1}{\csc x-\cot x}=\frac{1}{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}=\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}=...$

3. Ok

you should now that

cscx = 1/sinx
cotx = 1/tanx = cosx /sinx
sin^2x + cos^2x=1

1/(cscx - cotx ) = 1/( 1/sinx - cosx/sinx )
= 1/( (1-cosx)/sinx ) = sinx /(1-cosx)
multiply with 1+cosx/(1+cosx)
so you have

sinx (1+cosx)/((1-cosx)(1+cosx))
sinx(1+cosx)/(1-cos^2)
sinx(1+cosx)/sin^x.............since sin^2x + cos^2x = 1...sin^2x=1-cos^x
(1+cosx)/sinx

Edit: I was late

4. I appreciate the help from the both of you. I have to get used to thinking about how I can finagle these things into what I want.