Expand Using Sum/Difference Formulas cot(-(7pi)/12)

$\cot (- \frac{7 π}{12})$

Step 1

Replace $\cot (- \frac{7 π}{12})$ with an equivalent expression $\frac{1}{\tan (- \frac{7 π}{12})}$ using the fundamental identities.

$\frac{1}{\tan (- \frac{7 π}{12})}$

Step 2

Use a sum or difference formula on the denominator.

First, split the angle into two angles where the values of the six trigonometric functions are known. In this case, $- \frac{7 π}{12}$ can be split into $\frac{π}{4} - \frac{5 π}{6}$ .

$\frac{1}{\tan (\frac{π}{4} - \frac{5 π}{6})}$

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

$\frac{1}{\frac{\tan (\frac{π}{4}) - \tan (\frac{5 π}{6})}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

Remove parentheses.

$\frac{1}{\frac{\tan (\frac{π}{4}) - \tan (\frac{5 π}{6})}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

Simplify the numerator.

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$\frac{1}{\frac{1 - \tan (\frac{5 π}{6})}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

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.

$\frac{1}{\frac{1 + \tan (\frac{π}{6})}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

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

$\frac{1}{\frac{1 + \frac{\sqrt{3}}{3}}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

Multiply $- - \frac{\sqrt{3}}{3}$ .

Multiply $- 1$ by $- 1$ .

$\frac{1}{\frac{1 + 1 (\frac{\sqrt{3}}{3})}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

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

$\frac{1}{\frac{1 + \frac{\sqrt{3}}{3}}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

Write $1$ as a fraction with a common denominator.

$\frac{1}{\frac{\frac{3}{3} + \frac{\sqrt{3}}{3}}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

Combine the numerators over the common denominator.

$\frac{1}{\frac{\frac{3 + \sqrt{3}}{3}}{1 + \tan (\frac{π}{4}) \tan (\frac{5 π}{6})}}$

Simplify the denominator.

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$\frac{1}{\frac{\frac{3 + \sqrt{3}}{3}}{1 + 1 \tan (\frac{5 π}{6})}}$

Multiply $\tan (\frac{5 π}{6})$ by $1$ .

$\frac{1}{\frac{\frac{3 + \sqrt{3}}{3}}{1 + \tan (\frac{5 π}{6})}}$

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.

$\frac{1}{\frac{\frac{3 + \sqrt{3}}{3}}{1 - \tan (\frac{π}{6})}}$

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

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{1}{3 - \sqrt{3}}$ by $\frac{3 + \sqrt{3}}{3 + \sqrt{3}}$ .

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

Simplify terms.

Multiply $\frac{1}{3 - \sqrt{3}}$ by $\frac{3 + \sqrt{3}}{3 + \sqrt{3}}$ .

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

Expand the denominator using the FOIL method.

$\frac{1}{(3 + \sqrt{3}) (\frac{3 + \sqrt{3}}{9 + 3 \sqrt{3} - 3 \sqrt{3} - {\sqrt{3}}^{2}})}$

Simplify.

$\frac{1}{(3 + \sqrt{3}) (\frac{3 + \sqrt{3}}{6})}$

Apply the distributive property.

$\frac{1}{3 (\frac{3 + \sqrt{3}}{6}) + \sqrt{3} (\frac{3 + \sqrt{3}}{6})}$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$\frac{1}{3 (\frac{3 + \sqrt{3}}{3 (2)}) + \sqrt{3} (\frac{3 + \sqrt{3}}{6})}$

Cancel the common factor.

$\frac{1}{3 (\frac{3 + \sqrt{3}}{3 \cdot 2}) + \sqrt{3} (\frac{3 + \sqrt{3}}{6})}$

Rewrite the expression.

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \sqrt{3} (\frac{3 + \sqrt{3}}{6})}$

Combine $\sqrt{3}$ and $\frac{3 + \sqrt{3}}{6}$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{\sqrt{3} (3 + \sqrt{3})}{6}}$

Simplify each term.

Apply the distributive property.

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{\sqrt{3} \cdot 3 + \sqrt{3} \sqrt{3}}{6}}$

Move $3$ to the left of $\sqrt{3}$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 \cdot \sqrt{3} + \sqrt{3} \sqrt{3}}{6}}$

Combine using the product rule for radicals.

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 \cdot \sqrt{3} + \sqrt{3 \cdot 3}}{6}}$

Simplify each term.

Multiply $3$ by $3$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 \sqrt{3} + \sqrt{9}}{6}}$

Rewrite $9$ as $3^{2}$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 \sqrt{3} + \sqrt{3^{2}}}{6}}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 \sqrt{3} + 3}{6}}$

Cancel the common factor of $3 \sqrt{3} + 3$ and $6$ .

Factor $3$ out of $3 \sqrt{3}$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 (\sqrt{3}) + 3}{6}}$

Factor $3$ out of $3$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 (\sqrt{3}) + 3 (1)}{6}}$

Factor $3$ out of $3 (\sqrt{3}) + 3 (1)$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 (\sqrt{3} + 1)}{6}}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 (\sqrt{3} + 1)}{3 (2)}}$

Cancel the common factor.

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{3 (\sqrt{3} + 1)}{3 \cdot 2}}$

Rewrite the expression.

$\frac{1}{\frac{3 + \sqrt{3}}{2} + \frac{\sqrt{3} + 1}{2}}$

Simplify terms.

Combine the numerators over the common denominator.

$\frac{1}{\frac{3 + \sqrt{3} + \sqrt{3} + 1}{2}}$

Add $3$ and $1$ .

$\frac{1}{\frac{4 + \sqrt{3} + \sqrt{3}}{2}}$

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

$\frac{1}{\frac{4 + 2 \sqrt{3}}{2}}$

Cancel the common factor of $4 + 2 \sqrt{3}$ and $2$ .

Factor $2$ out of $4$ .

$\frac{1}{\frac{2 \cdot 2 + 2 \sqrt{3}}{2}}$

Factor $2$ out of $2 \sqrt{3}$ .

$\frac{1}{\frac{2 \cdot 2 + 2 (\sqrt{3})}{2}}$

Factor $2$ out of $2 (2) + 2 (\sqrt{3})$ .

$\frac{1}{\frac{2 (2 + \sqrt{3})}{2}}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{1}{\frac{2 (2 + \sqrt{3})}{2 (1)}}$

Cancel the common factor.

$\frac{1}{\frac{2 (2 + \sqrt{3})}{2 \cdot 1}}$

Rewrite the expression.

$\frac{1}{\frac{2 + \sqrt{3}}{1}}$

Divide $2 + \sqrt{3}$ by $1$ .

$\frac{1}{2 + \sqrt{3}}$

Step 3

Simplify.

Multiply $\frac{1}{2 + \sqrt{3}}$ by $\frac{2 - \sqrt{3}}{2 - \sqrt{3}}$ .

$\frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$

Multiply $\frac{1}{2 + \sqrt{3}}$ by $\frac{2 - \sqrt{3}}{2 - \sqrt{3}}$ .

$\frac{2 - \sqrt{3}}{(2 + \sqrt{3}) (2 - \sqrt{3})}$

Expand the denominator using the FOIL method.

$\frac{2 - \sqrt{3}}{4 - 2 \sqrt{3} + \sqrt{3} \cdot 2 - {\sqrt{3}}^{2}}$

Simplify.

$\frac{2 - \sqrt{3}}{1}$

Divide $2 - \sqrt{3}$ by $1$ .

$2 - \sqrt{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$2 - \sqrt{3}$

Decimal Form:

$0.26794919 \dots$

Expand Using Sum/Difference Formulas cot(-(7pi)/12)

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