1. Proving a trigonometric identity

I came across this in the calculus section: $\dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. How can it be proven?

2. Originally Posted by PQR
I came across this in the calculus section: $\dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. How can it be proven?
Hint: $1-\cos(2x) = 1-(2\cos^2x-1) = 2(1-\cos^2x)$

3. Originally Posted by PQR
I came across this in the calculus section: $\dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. How can it be proven?
L.H.S

Edited:someone give you a hint

4. Originally Posted by Chris L T521
Hint: $1-\cos(2x) = 1-(2\cos^2x-1) = 2(1-\cos^2x)$
$\therefore \dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{2\left(\cos{x}-1\right)^2}{2\left(1-\cos^2{x}\right)}$ $= \dfrac{\left(\cos{x}-1\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(1+\cos{x}\right)}$ $= \dfrac{\left(1-\cos{x}\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(-1-\cos{x}\right)} = \dfrac{\left(\cos{x}-1\right)}{\left(-1-\cos{x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$.

5. Originally Posted by PQR
$\therefore \dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{2\left(\cos{x}-1\right)^2}{2\left(1-\cos^2{x}\right)}$ $= \dfrac{\left(\cos{x}-1\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(1+\cos{x}\right)}$ $= \dfrac{\left(1-\cos{x}\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(-1-\cos{x}\right)} = \dfrac{\left(\cos{x}-1\right)}{\left(-1-\cos{x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$.
Good job!