# How do you prove #\frac { \cos A + \sin A } { \cos A - \sin A } = \tan ( 45^ { \circ } + A )#?

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To prove ( \frac{\cos A + \sin A}{\cos A - \sin A} = \tan(45^\circ + A) ), we can start with the right-hand side and express it in terms of sine and cosine using trigonometric identities. Then, simplify and manipulate both sides until they are equal.

Using the angle addition formula for tangent, we have:

[ \tan(45^\circ + A) = \frac{\sin(45^\circ + A)}{\cos(45^\circ + A)} ]

Applying the angle addition formulas for sine and cosine:

[ \sin(45^\circ + A) = \sin 45^\circ \cos A + \cos 45^\circ \sin A = \frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A ]

[ \cos(45^\circ + A) = \cos 45^\circ \cos A - \sin 45^\circ \sin A = \frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A ]

Substitute these expressions into the tangent expression:

[ \tan(45^\circ + A) = \frac{\frac{1}{\sqrt{2}} \cos A + \frac{1}{\sqrt{2}} \sin A}{\frac{1}{\sqrt{2}} \cos A - \frac{1}{\sqrt{2}} \sin A} ]

[ = \frac{\frac{1}{\sqrt{2}}(\cos A + \sin A)}{\frac{1}{\sqrt{2}}(\cos A - \sin A)} ]

[ = \frac{\cos A + \sin A}{\cos A - \sin A} ]

Thus, we have shown that ( \frac{\cos A + \sin A}{\cos A - \sin A} = \tan(45^\circ + A) ).

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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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