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

Verify

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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 ).

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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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