# How do you simplify #tanx + cosx/(1 + sinx)#?

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To simplify ( \frac{\tan(x) + \cos(x)}{1 + \sin(x)} ), we can use trigonometric identities to rewrite the expression.

- First, recall that ( \tan(x) = \frac{\sin(x)}{\cos(x)} ).
- Next, we know that ( \cos(x) = \frac{1}{\sec(x)} ) and ( \sin(x) = \frac{1}{\csc(x)} ).

Substitute these into the expression:

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

Now, simplify the expression:

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

So, ( \frac{\tan(x) + \cos(x)}{1 + \sin(x)} ) simplifies to ( \frac{1}{1 + \sin(x)} ).

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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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