How do you prove that #costheta/(1sintheta) + costheta/(1+sintheta) = 2 /costheta#?
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To prove ( \frac{\cos(\theta)}{1\sin(\theta)} + \frac{\cos(\theta)}{1+\sin(\theta)} = \frac{2}{\cos(\theta)} ), follow these steps:

Start with the lefthand side of the equation: [ \frac{\cos(\theta)}{1\sin(\theta)} + \frac{\cos(\theta)}{1+\sin(\theta)} ]

Find a common denominator for the fractions: [ \frac{\cos(\theta)(1+\sin(\theta))}{(1\sin(\theta))(1+\sin(\theta))} + \frac{\cos(\theta)(1\sin(\theta))}{(1\sin(\theta))(1+\sin(\theta))} ]

Combine the fractions: [ \frac{\cos(\theta)(1+\sin(\theta)) + \cos(\theta)(1\sin(\theta))}{(1\sin(\theta))(1+\sin(\theta))} ]

Simplify the numerator: [ \cos(\theta) + \cos(\theta)\sin(\theta) + \cos(\theta)  \cos(\theta)\sin(\theta) ]

Combine like terms in the numerator: [ 2\cos(\theta) ]

Substitute the simplified numerator back into the equation: [ \frac{2\cos(\theta)}{(1\sin(\theta))(1+\sin(\theta))} ]

Use the identity ( 1  \sin^2(\theta) = \cos^2(\theta) ): [ \frac{2\cos(\theta)}{\cos^2(\theta)} ]

Simplify: [ \frac{2}{\cos(\theta)} ]
This is the righthand side of the equation, thus proving the identity.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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