# How do you prove #(1 - tanx) / (1 + tanx) = (1 - sin2x) / (cos2x)#?

To find the solution, multiply the left side of the denominator and numerator by 1-tanx and then simplify:

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To prove ( \frac{1 - \tan(x)}{1 + \tan(x)} = \frac{1 - \sin(2x)}{\cos(2x)} ), we can start with the left side of the equation and manipulate it using trigonometric identities.

Starting with the left side:

[ \frac{1 - \tan(x)}{1 + \tan(x)} ]

We can rewrite ( \tan(x) ) in terms of sine and cosine using the identity ( \tan(x) = \frac{\sin(x)}{\cos(x)} ):

[ = \frac{1 - \frac{\sin(x)}{\cos(x)}}{1 + \frac{\sin(x)}{\cos(x)}} ]

Now, to simplify this expression, we can find a common denominator:

[ = \frac{\cos(x) - \sin(x)}{\cos(x) + \sin(x)} ]

Next, we can multiply both the numerator and denominator by ( \cos(x) ) to eliminate the fraction:

[ = \frac{\cos(x)(\cos(x) - \sin(x))}{\cos(x)(\cos(x) + \sin(x))} ]

Expanding the numerator and denominator:

[ = \frac{\cos^2(x) - \cos(x)\sin(x)}{\cos^2(x) + \cos(x)\sin(x)} ]

Using the double-angle identity for sine, ( \sin(2x) = 2\sin(x)\cos(x) ), and the Pythagorean identity for cosine, ( \cos^2(x) = 1 - \sin^2(x) ):

[ = \frac{1 - \sin^2(x) - \sin(x)\cos(x)}{1 - \sin^2(x) + \sin(x)\cos(x)} ]

Now, factor out ( 1 - \sin^2(x) ) from both the numerator and the denominator:

[ = \frac{(1 - \sin^2(x))(1 - \sin(x))}{(1 - \sin^2(x))(1 + \sin(x))} ]

Cancel out the common factor of ( 1 - \sin^2(x) ):

[ = \frac{1 - \sin(x)}{1 + \sin(x)} ]

Finally, using the double-angle identity for sine, ( \sin(2x) = 2\sin(x)\cos(x) ), and the Pythagorean identity for cosine, ( \cos(2x) = 1 - 2\sin^2(x) ):

[ = \frac{1 - \sin(2x)}{\cos(2x)} ]

This matches the right side of the equation, proving the identity.

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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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