How do you verify #(1 - cos x) / (1 + cos x) = (csc x - cot x)^2#?

Question

Madelyn Adams · Accepted Answer

Verify #(1 - cos x)/(1 + cos x) = (csc x - cot x)^2# Multiply both numerator and denominator of left side by (1 - cos x). #((1 - cos x)/(1 + cos x))((1 - cos x)/(1 - cos x)) =# #= (1 - cos x)^2/(1 - cos^2 x) = (1 - cos x)^2/(sin^2 x) = # #= ((1- cos x)/(sin x))^2 = (1/(sin x) - (cos x/sin x))^2 = # #= (csc x - cot x)^2#

Oliver Atkinson · Answer

undefined To verify $ \frac{{1 - \cos x}}{{1 + \cos x}} = (\csc x - \cot x)^2 $, we can start with the left-hand side (LHS) of the equation and manipulate it until it matches the right-hand side (RHS).

Starting with the LHS:
$$ \frac{{1 - \cos x}}{{1 + \cos x}} $$

We can multiply both the numerator and the denominator by $ (1 - \cos x) $ to rationalize the expression:
$$ = \frac{{(1 - \cos x)(1 - \cos x)}}{{(1 + \cos x)(1 - \cos x)}} $$

Expanding the numerator:
$$ = \frac{{1 - 2\cos x + \cos^2 x}}{{1 - \cos^2 x}} $$

Using the Pythagorean identity $ \sin^2 x + \cos^2 x = 1 $:
$$ = \frac{{1 - 2\cos x + (1 - \sin^2 x)}}{{(1 + \cos x)(1 - \cos x)}} $$

Simplifying:
$$ = \frac{{2 - 2\cos x - \sin^2 x}}{{(1 + \cos x)(1 - \cos x)}} $$

Using the Pythagorean identity $ \sin^2 x = 1 - \cos^2 x $:
$$ = \frac{{2 - 2\cos x - (1 - \cos^2 x)}}{{(1 + \cos x)(1 - \cos x)}} $$

$$ = \frac{{2 - 2\cos x - 1 + \cos^2 x}}{{(1 + \cos x)(1 - \cos x)}} $$

$$ = \frac{{1 - 2\cos x + \cos^2 x}}{{(1 + \cos x)(1 - \cos x)}} $$

Now, the expression in the numerator matches $ (\csc x - \cot x)^2 $. Hence, we have verified that $ \frac{{1 - \cos x}}{{1 + \cos x}} = (\csc x - \cot x)^2 $.