# Prove 1+cos x/1-cos x = 1+ 2cos x(1+cos x)/sinˆ2 Please and thank you! (?)

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The picture is just in case it wasn't clearly typed. It's question 23. Steps would be greatly appreciated.

We are using these to help us but I forget the steps.

The picture is just in case it wasn't clearly typed. It's question 23. Steps would be greatly appreciated.

We are using these to help us but I forget the steps.

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

Starting with the left-hand side (LHS): [ \frac{1 + \cos x}{1 - \cos x} ] [ = \frac{(1 + \cos x)(1 + \cos x)}{(1 - \cos x)(1 + \cos x)} ] [ = \frac{(1 + \cos x)^2}{1 - \cos^2 x} ] [ = \frac{(1 + \cos x)^2}{\sin^2 x} ] (using the identity (1 - \cos^2 x = \sin^2 x))

Now, for the right-hand side (RHS): [ \frac{1 + 2\cos x(1 + \cos x)}{\sin^2 x} ] [ = \frac{1 + 2\cos x + 2\cos^2 x}{\sin^2 x} ] [ = \frac{(1 + \cos x)^2}{\sin^2 x} ]

Thus, the LHS is equal to the RHS. Hence, the given identity is proved.

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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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