How do you prove the identity #1 / (1- cosx) + 1/ (1+ cosx) = 2 csc^2x#?

Question

Samantha Abate · Accepted Answer

Utilizing the left side of the body #((1+cosx)+(1-cosx))/((1-cosx)(1+cosx))# #(1+cosx+1-cosx)/(1-cos^2x)# #2/sin^2x# #2csc^2x# which corresponds to the right side

Elijah Alexander · Answer

undefined To prove the identity $ \frac{1}{1 - \cos(x)} + \frac{1}{1 + \cos(x)} = 2 \csc^2(x) $, you can use trigonometric identities such as the reciprocal identity $ \csc(x) = \frac{1}{\sin(x)} $ and the Pythagorean identity $ \sin^2(x) + \cos^2(x) = 1 $. Here's the proof:

Starting with the left side:
$$ \frac{1}{1 - \cos(x)} + \frac{1}{1 + \cos(x)} $$

Finding a common denominator:
$$ = \frac{1 + \cos(x)}{(1 - \cos(x))(1 + \cos(x))} + \frac{1 - \cos(x)}{(1 - \cos(x))(1 + \cos(x))} $$

Combining the fractions:
$$ = \frac{(1 + \cos(x)) + (1 - \cos(x))}{1 - \cos^2(x)} $$

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

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

Recalling the reciprocal identity $ \csc(x) = \frac{1}{\sin(x)} $:
$$ = 2 \csc^2(x) $$

Thus, the left side is equivalent to $ 2 \csc^2(x) $, which matches the right side. Therefore, the identity is proven.