Trigonometry Examples

$\tan (120)$

Step 1

The angle $120$ is an angle where the values of the six trigonometric functions are known. Because this is the case, add $0$ to keep the value the same.

$\tan (120 + 0)$

Step 2

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

$\frac{\tan (120) + \tan (0)}{1 - \tan (120) \tan (0)}$

Step 3

Simplify the numerator.

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Step 3.1

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.

$\frac{- \tan (60) + \tan (0)}{1 - \tan (120) \tan (0)}$

Step 3.2

The exact value of $\tan (60)$ is $\sqrt{3}$ .

$\frac{- \sqrt{3} + \tan (0)}{1 - \tan (120) \tan (0)}$

Step 3.3

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

$\frac{- \sqrt{3} + 0}{1 - \tan (120) \tan (0)}$

Step 3.4

Add $- \sqrt{3}$ and $0$ .

$\frac{- \sqrt{3}}{1 - \tan (120) \tan (0)}$

Step 4

Simplify the denominator.

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Step 4.1

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.

$\frac{- \sqrt{3}}{1 - - \tan (60) \tan (0)}$

Step 4.2

The exact value of $\tan (60)$ is $\sqrt{3}$ .

$\frac{- \sqrt{3}}{1 - - \sqrt{3} \tan (0)}$

Step 4.3

Multiply $- - \sqrt{3}$ .

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Step 4.3.1

Multiply $- 1$ by $- 1$ .

$\frac{- \sqrt{3}}{1 + 1 \sqrt{3} \tan (0)}$

Step 4.3.2

Multiply $\sqrt{3}$ by $1$ .

$\frac{- \sqrt{3}}{1 + \sqrt{3} \tan (0)}$

Step 4.4

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

$\frac{- \sqrt{3}}{1 + \sqrt{3} \cdot 0}$

Step 4.5

Multiply $\sqrt{3}$ by $0$ .

$\frac{- \sqrt{3}}{1 + 0}$

Step 4.6

Add $1$ and $0$ .

$\frac{- \sqrt{3}}{1}$

Step 5

Divide $- \sqrt{3}$ by $1$ .

$- \sqrt{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$- \sqrt{3}$

Decimal Form:

$- 1.73205080 \dots$