Expand Using Sum/Difference Formulas tan(pi+theta)

$\tan (π + θ)$

Use the sum formula for tangent to simplify the expression. The formula states that $\tan (A + B) = \frac{\tan (A) + \tan (B)}{1 - \tan (A) \tan (B)}$ .

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

Simplify the numerator.

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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.

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

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

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

Multiply $- 1$ by $0$ .

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

Add $0$ and $\tan (θ)$ .

$\frac{\tan (θ)}{1 - \tan (π) \tan (θ)}$

Simplify the denominator.

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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.

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

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

$\frac{\tan (θ)}{1 - - 0 \tan (θ)}$

Multiply $- 1$ by $0$ .

$\frac{\tan (θ)}{1 - 0 \tan (θ)}$

Multiply $- 1$ by $0$ .

$\frac{\tan (θ)}{1 + 0 \tan (θ)}$

Multiply $0$ by $\tan (θ)$ .

$\frac{\tan (θ)}{1 + 0}$

Add $1$ and $0$ .

$\frac{\tan (θ)}{1}$

Divide $\tan (θ)$ by $1$ .

$\tan (θ)$

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