# How do I prove that #1/(sec A+1)+1/(sec A-1)#=#2 csc A cot A# ?

The Lowest Common Multiple is taken.

Changing,

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To prove that (\frac{1}{(\sec A + 1)} + \frac{1}{(\sec A - 1)} = 2 \csc A \cot A), we can start with the left side of the equation and manipulate it until we reach the expression on the right side.

Starting with the left side:

[ \frac{1}{(\sec A + 1)} + \frac{1}{(\sec A - 1)} ]

We'll first find a common denominator:

[ \frac{(\sec A - 1) + (\sec A + 1)}{(\sec A + 1)(\sec A - 1)} ]

Combining the fractions:

[ \frac{(\sec A - 1) + (\sec A + 1)}{(\sec A + 1)(\sec A - 1)} = \frac{2\sec A}{\sec^2 A - 1} ]

Using the Pythagorean identity (\sec^2 A - 1 = \tan^2 A):

[ \frac{2\sec A}{\sec^2 A - 1} = \frac{2\sec A}{\tan^2 A} ]

Using the reciprocal identities (\sec A = \frac{1}{\cos A}) and (\tan A = \frac{\sin A}{\cos A}):

[ \frac{2\sec A}{\tan^2 A} = \frac{2(\frac{1}{\cos A})}{(\frac{\sin A}{\cos A})^2} ]

Simplify:

[ \frac{2(\frac{1}{\cos A})}{(\frac{\sin A}{\cos A})^2} = \frac{2}{\frac{\sin^2 A}{\cos A}} ]

[ \frac{2}{\frac{\sin^2 A}{\cos A}} = 2 \csc A \cot A ]

Thus, we have proven that (\frac{1}{(\sec A + 1)} + \frac{1}{(\sec A - 1)} = 2 \csc A \cot 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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