Trigonometry Examples

\tan (15 π - 2 t)

$\tan (15 π - 2 t)$

Step 1

Use the difference formula for tangent to simplify the expression. The formula states that

\tan (A - B) = \frac{\tan (A) - \tan (B)}{1 + \tan (A) \tan (B)}

$\tan (A - B) = \frac{\tan (A) - \tan (B)}{1 + \tan (A) \tan (B)}$ .

\frac{\tan (15 π) - \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}

$\frac{\tan (15 π) - \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}$

Step 2

Remove parentheses.

\frac{\tan (15 π) - \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}

$\frac{\tan (15 π) - \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}$

Step 3

Simplify the numerator.

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Step 3.1

Factor

- 1

$- 1$ out of

\tan (15 π) - \tan (2 t)

$\tan (15 π) - \tan (2 t)$ .

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Step 3.1.1

Factor

- 1

$- 1$ out of

\tan (15 π)

$\tan (15 π)$ .

\frac{- 1 (- \tan (15 π)) - \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}

$\frac{- 1 (- \tan (15 π)) - \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}$

Step 3.1.2

Factor

- 1

$- 1$ out of

- \tan (2 t)

$- \tan (2 t)$ .

\frac{- 1 (- \tan (15 π)) - (\tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- 1 (- \tan (15 π)) - (\tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.1.3

Factor

- 1

$- 1$ out of

- 1 (- \tan (15 π)) - (\tan (2 t))

$- 1 (- \tan (15 π)) - (\tan (2 t))$ .

\frac{- 1 (- \tan (15 π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- 1 (- \tan (15 π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.1.4

Rewrite

- 1 (- \tan (15 π) + \tan (2 t))

$- 1 (- \tan (15 π) + \tan (2 t))$ as

- (- \tan (15 π) + \tan (2 t))

$- (- \tan (15 π) + \tan (2 t))$ .

\frac{- (- \tan (15 π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (- \tan (15 π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

\frac{- (- \tan (15 π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (- \tan (15 π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.2

Subtract full rotations of

2 π

$2 π$ until the angle is greater than or equal to

0

$0$ and less than

2 π

$2 π$ .

\frac{- (- \tan (π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (- \tan (π) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.3

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 (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (- - \tan (0) + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.4

The exact value of

\tan (0)

$\tan (0)$ is

0

$0$ .

\frac{- (- - 0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (- - 0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.5

Multiply

- - 0

$- - 0$ .

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Step 3.5.1

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{- (- 0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (- 0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.5.2

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{- (0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

\frac{- (0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}

$\frac{- (0 + \tan (2 t))}{1 + \tan (15 π) \tan (2 t)}$

Step 3.6

Add

0

$0$ and

\tan (2 t)

$\tan (2 t)$ .

\frac{- \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}

$\frac{- \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}$

\frac{- \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}

$\frac{- \tan (2 t)}{1 + \tan (15 π) \tan (2 t)}$

Step 4

Simplify the denominator.

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Step 4.1

Subtract full rotations of

2 π

$2 π$ until the angle is greater than or equal to

0

$0$ and less than

2 π

$2 π$ .

\frac{- \tan (2 t)}{1 + \tan (π) \tan (2 t)}

$\frac{- \tan (2 t)}{1 + \tan (π) \tan (2 t)}$

Step 4.2

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 (2 t)}{1 - \tan (0) \tan (2 t)}

$\frac{- \tan (2 t)}{1 - \tan (0) \tan (2 t)}$

Step 4.3

The exact value of

\tan (0)

$\tan (0)$ is

0

$0$ .

\frac{- \tan (2 t)}{1 - 0 \tan (2 t)}

$\frac{- \tan (2 t)}{1 - 0 \tan (2 t)}$

Step 4.4

Multiply

- 1

$- 1$ by

0

$0$ .

\frac{- \tan (2 t)}{1 + 0 \tan (2 t)}

$\frac{- \tan (2 t)}{1 + 0 \tan (2 t)}$

Step 4.5

Multiply

0

$0$ by

\tan (2 t)

$\tan (2 t)$ .

\frac{- \tan (2 t)}{1 + 0}

$\frac{- \tan (2 t)}{1 + 0}$

Step 4.6

Add

1

$1$ and

0

$0$ .

\frac{- \tan (2 t)}{1}

$\frac{- \tan (2 t)}{1}$

\frac{- \tan (2 t)}{1}

$\frac{- \tan (2 t)}{1}$

Step 5

Divide

- \tan (2 t)

$- \tan (2 t)$ by

1

$1$ .

- \tan (2 t)

$- \tan (2 t)$