# Trig Derivative

• Oct 11th 2009, 07:04 PM
VitaX
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.

$r' = 2 \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right) \left(\frac{(1-Cos \Theta)(Cos \Theta) - (1 + Sin \Theta)(Sin \Theta)}{(1-Cos \Theta)^2}\right)$
• Oct 11th 2009, 07:26 PM
Arturo_026
looks good to me, from there on is pure trig.
• Oct 11th 2009, 07:34 PM
VitaX
$r' = 2 \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right) \left(\frac{Cos \Theta - Cos^2 \Theta - Sin \Theta + Sin^2 \Theta}{(1-Cos \Theta)^2}\right)$