Trigonometry Examples

$\tan (\frac{- 5 π}{6}) - \cot (\frac{7 π}{2})$

Step 1

Simplify each term.

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Step 1.1

Move the negative in front of the fraction.

$\tan (- \frac{5 π}{6}) - \cot (\frac{7 π}{2})$

Step 1.2

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\tan (\frac{7 π}{6}) - \cot (\frac{7 π}{2})$

Step 1.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$\tan (\frac{π}{6}) - \cot (\frac{7 π}{2})$

Step 1.4

The exact value of $\tan (\frac{π}{6})$ is $\frac{\sqrt{3}}{3}$ .

$\frac{\sqrt{3}}{3} - \cot (\frac{7 π}{2})$

Step 1.5

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{\sqrt{3}}{3} - \cot (\frac{3 π}{2})$

Step 1.6

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

$\frac{\sqrt{3}}{3} - - \cot (\frac{π}{2})$

Step 1.7

The exact value of $\cot (\frac{π}{2})$ is $0$ .

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

Step 1.8

Multiply $- - 0$ .

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Step 1.8.1

Multiply $- 1$ by $0$ .

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

Step 1.8.2

Multiply $- 1$ by $0$ .

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

Step 2

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

$\frac{\sqrt{3}}{3}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{3}}{3}$

Decimal Form:

$0.57735026 \dots$