Precalculus Examples

$\frac{\tan (x) - \sin (- x)}{1 + \cos (x)} = \tan (x)$

Step 1

Start on the left side.

$\frac{\tan (x) - \sin (- x)}{1 + \cos (x)}$

Step 2

Since $\sin (- x)$ is an odd function, rewrite $\sin (- x)$ as $- \sin (x)$ .

$\frac{\tan (x) - - \sin (x)}{1 + \cos (x)}$

Step 3

Write $\tan (x)$ in sines and cosines using the quotient identity.

$\frac{\frac{\sin (x)}{\cos (x)} - - \sin (x)}{1 + \cos (x)}$

Step 4

Simplify.

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Step 4.1

Multiply the numerator and denominator of the fraction by $\cos (x)$ .

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Step 4.1.1

Multiply $\frac{\frac{\sin (x)}{\cos (x)} - - \sin (x)}{1 + \cos (x)}$ by $\frac{\cos (x)}{\cos (x)}$ .

$\frac{\cos (x)}{\cos (x)} \cdot \frac{\frac{\sin (x)}{\cos (x)} - - \sin (x)}{1 + \cos (x)}$

Step 4.1.2

Combine.

$\frac{\cos (x) (\frac{\sin (x)}{\cos (x)} - - \sin (x))}{\cos (x) (1 + \cos (x))}$

Step 4.2

Apply the distributive property.

$\frac{\cos (x) \frac{\sin (x)}{\cos (x)} + \cos (x) (- - \sin (x))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.3

Cancel the common factor of $\cos (x)$ .

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Step 4.3.1

Cancel the common factor.

$\frac{\cos (x) \frac{\sin (x)}{\cos (x)} + \cos (x) (- - \sin (x))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.3.2

Rewrite the expression.

$\frac{\sin (x) + \cos (x) (- - \sin (x))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.4

Simplify the numerator.

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Step 4.4.1

Factor $\sin (x)$ out of $\sin (x) + \cos (x) (- - \sin (x))$ .

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Step 4.4.1.1

Multiply by $1$ .

$\frac{\sin (x) \cdot 1 + \cos (x) (- - \sin (x))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.4.1.2

Factor $\sin (x)$ out of $\cos (x) (- - \sin (x))$ .

$\frac{\sin (x) \cdot 1 + \sin (x) (\cos (x) (- - 1))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.4.1.3

Factor $\sin (x)$ out of $\sin (x) \cdot 1 + \sin (x) (\cos (x) (- - 1))$ .

$\frac{\sin (x) (1 + \cos (x) (- - 1))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.4.2

Multiply $- 1$ by $- 1$ .

$\frac{\sin (x) (1 + \cos (x) \cdot 1)}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.4.3

Multiply $\cos (x)$ by $1$ .

$\frac{\sin (x) (1 + \cos (x))}{\cos (x) \cdot 1 + \cos (x) \cos (x)}$

Step 4.5

Factor $\cos (x)$ out of $\cos (x) \cdot 1 + \cos (x) \cos (x)$ .

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Step 4.5.1

Factor $\cos (x)$ out of $\cos (x) \cdot 1$ .

$\frac{\sin (x) (1 + \cos (x))}{\cos (x) (1) + \cos (x) \cos (x)}$

Step 4.5.2

Factor $\cos (x)$ out of $\cos (x) \cos (x)$ .

$\frac{\sin (x) (1 + \cos (x))}{\cos (x) (1) + \cos (x) (\cos (x))}$

Step 4.5.3

Factor $\cos (x)$ out of $\cos (x) (1) + \cos (x) (\cos (x))$ .

$\frac{\sin (x) (1 + \cos (x))}{\cos (x) (1 + \cos (x))}$

Step 4.6

Cancel the common factor of $1 + \cos (x)$ .

$\frac{\sin (x)}{\cos (x)}$

Step 5

Rewrite $\frac{\sin (x)}{\cos (x)}$ as $\tan (x)$ .

$\tan (x)$

Step 6

Because the two sides have been shown to be equivalent, the equation is an identity.

$\frac{\tan (x) - \sin (- x)}{1 + \cos (x)} = \tan (x)$ is an identity