Trigonometry Examples

$\tan (π - x) = - \tan (x)$

Step 1

Start on the left side.

$\tan (π - x)$

Step 2

Apply the difference of angles identity.

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Factor $- 1$ out of $\tan (π) - \tan (x)$ .

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Step 3.1.1.1

Factor $- 1$ out of $\tan (π)$ .

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

Step 3.1.1.2

Factor $- 1$ out of $- \tan (x)$ .

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

Step 3.1.1.3

Factor $- 1$ out of $- 1 (- \tan (π)) - (\tan (x))$ .

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

Step 3.1.1.4

Rewrite $- 1 (- \tan (π) + \tan (x))$ as $- (- \tan (π) + \tan (x))$ .

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

Step 3.1.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$\frac{- (- - \tan (0) + \tan (x))}{1 + \tan (π) \tan (x)}$

Step 3.1.3

The exact value of $\tan (0)$ is $0$ .

$\frac{- (- - 0 + \tan (x))}{1 + \tan (π) \tan (x)}$

Step 3.1.4

Multiply $- - 0$ .

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Step 3.1.4.1

Multiply $- 1$ by $0$ .

$\frac{- (- 0 + \tan (x))}{1 + \tan (π) \tan (x)}$

Step 3.1.4.2

Multiply $- 1$ by $0$ .

$\frac{- (0 + \tan (x))}{1 + \tan (π) \tan (x)}$

Step 3.1.5

Add $0$ and $\tan (x)$ .

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

Step 3.2

Simplify the denominator.

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Step 3.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$\frac{- \tan (x)}{1 - \tan (0) \tan (x)}$

Step 3.2.2

The exact value of $\tan (0)$ is $0$ .

$\frac{- \tan (x)}{1 - 0 \tan (x)}$

Step 3.2.3

Multiply $- 1$ by $0$ .

$\frac{- \tan (x)}{1 + 0 \tan (x)}$

Step 3.2.4

Multiply $0$ by $\tan (x)$ .

$\frac{- \tan (x)}{1 + 0}$

Step 3.2.5

Add $1$ and $0$ .

$\frac{- \tan (x)}{1}$

Step 3.3

Divide $- \tan (x)$ by $1$ .

$- \tan (x)$

Step 4

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

$\tan (π - x) = - \tan (x)$ is an identity