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

Utilizing the left side of the body

which corresponds to the right side

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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