# Thread: Trig Derivative

1. ## Trig Derivative

Find the derivative

$r = \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right)^2$

I believe the chain rule and quotient rule are used here. So here's my first step, if someone can confirm I did this correct than I could probably go on from there.

$r' = 2 \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right) \left(\frac{(1-Cos \Theta)(Cos \Theta) - (1 + Sin \Theta)(Sin \Theta)}{(1-Cos \Theta)^2}\right)$

2. looks good to me, from there on is pure trig.

3. $r' = 2 \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right) \left(\frac{Cos \Theta - Cos^2 \Theta - Sin \Theta + Sin^2 \Theta}{(1-Cos \Theta)^2}\right)$

Would be final answer right?

4. If you want to complicate it

5. Originally Posted by Arturo_026

If you want to complicate it
Nah I'll stick with what I posted xD