Trigonometry Examples

$\frac{2 \tan (\frac{7 π}{12})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1

Simplify the numerator.

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Step 1.1

The exact value of $\tan (\frac{7 π}{12})$ is $- \sqrt{7 + 4 \sqrt{3}}$ .

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Step 1.1.1

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\frac{2 \tan (\frac{\frac{7 π}{6}}{2})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.2

Apply the tangent half-angle identity.

$\frac{2 (\pm \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.3

Change the $\pm$ to $-$ because tangent is negative in the second quadrant.

$\frac{2 (- \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4

Simplify $- \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}}$ .

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Step 1.1.4.1

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

$\frac{2 (- \sqrt{\frac{1 - - \cos (\frac{π}{6})}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{2 (- \sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.3

Multiply $- - \frac{\sqrt{3}}{2}$ .

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Step 1.1.4.3.1

Multiply $- 1$ by $- 1$ .

$\frac{2 (- \sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.3.2

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

$\frac{2 (- \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.4

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

$\frac{2 (- \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.5

Combine the numerators over the common denominator.

$\frac{2 (- \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.6

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

$\frac{2 (- \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \cos (\frac{π}{6})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.7

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{2 (- \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.8

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

$\frac{2 (- \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2}{2} - \frac{\sqrt{3}}{2}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.9

Combine the numerators over the common denominator.

$\frac{2 (- \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.10

Multiply the numerator by the reciprocal of the denominator.

$\frac{2 (- \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{2}{2 - \sqrt{3}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.11

Cancel the common factor of $2$ .

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Step 1.1.4.11.1

Cancel the common factor.

$\frac{2 (- \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{2}{2 - \sqrt{3}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.11.2

Rewrite the expression.

$\frac{2 (- \sqrt{(2 + \sqrt{3}) \frac{1}{2 - \sqrt{3}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.12

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

$\frac{2 (- \sqrt{(2 + \sqrt{3}) (\frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}})})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.13

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

$\frac{2 (- \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{(2 - \sqrt{3}) (2 + \sqrt{3})}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.14

Expand the denominator using the FOIL method.

$\frac{2 (- \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{4 + 2 \sqrt{3} - 2 \sqrt{3} - {\sqrt{3}}^{2}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.15

Simplify.

$\frac{2 (- \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{1}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.16

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

$\frac{2 (- \sqrt{(2 + \sqrt{3}) (2 + \sqrt{3})})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.17

Expand $(2 + \sqrt{3}) (2 + \sqrt{3})$ using the FOIL Method.

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Step 1.1.4.17.1

Apply the distributive property.

$\frac{2 (- \sqrt{2 (2 + \sqrt{3}) + \sqrt{3} (2 + \sqrt{3})})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.17.2

Apply the distributive property.

$\frac{2 (- \sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} (2 + \sqrt{3})})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.17.3

Apply the distributive property.

$\frac{2 (- \sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18

Simplify and combine like terms.

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Step 1.1.4.18.1

Simplify each term.

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Step 1.1.4.18.1.1

Multiply $2$ by $2$ .

$\frac{2 (- \sqrt{4 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.1.2

Move $2$ to the left of $\sqrt{3}$ .

$\frac{2 (- \sqrt{4 + 2 \sqrt{3} + 2 \cdot \sqrt{3} + \sqrt{3} \sqrt{3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.1.3

Combine using the product rule for radicals.

$\frac{2 (- \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3 \cdot 3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.1.4

Multiply $3$ by $3$ .

$\frac{2 (- \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{9}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.1.5

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

$\frac{2 (- \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3^{2}}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.1.6

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

$\frac{2 (- \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + 3})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.2

Add $4$ and $3$ .

$\frac{2 (- \sqrt{7 + 2 \sqrt{3} + 2 \sqrt{3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.1.4.18.3

Add $2 \sqrt{3}$ and $2 \sqrt{3}$ .

$\frac{2 (- \sqrt{7 + 4 \sqrt{3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

$\frac{2 \cdot - 1 \sqrt{7 + 4 \sqrt{3}}}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.2

Combine exponents.

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Step 1.2.1

Factor out negative.

$\frac{- (2 \sqrt{7 + 4 \sqrt{3}})}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 1.2.2

Multiply $2$ by $- 1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 - \tan^{2} (\frac{7 π}{12})}$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $1$ as $1^{2}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1^{2} - \tan^{2} (\frac{7 π}{12})}$

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \tan (\frac{7 π}{12})$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 + \tan (\frac{7 π}{12})) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3

Simplify.

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Step 2.3.1

The exact value of $\tan (\frac{7 π}{12})$ is $- \sqrt{7 + 4 \sqrt{3}}$ .

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Step 2.3.1.1

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 + \tan (\frac{\frac{7 π}{6}}{2})) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.2

Apply the tangent half-angle identity.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 \pm \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.3

Change the $\pm$ to $-$ because tangent is negative in the second quadrant.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4

Simplify $- \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}}$ .

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Step 2.3.1.4.1

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{1 - - \cos (\frac{π}{6})}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.3

Multiply $- - \frac{\sqrt{3}}{2}$ .

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Step 2.3.1.4.3.1

Multiply $- 1$ by $- 1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.3.2

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.4

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.5

Combine the numerators over the common denominator.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.6

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \cos (\frac{π}{6})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.7

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.8

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2}{2} - \frac{\sqrt{3}}{2}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.9

Combine the numerators over the common denominator.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.10

Multiply the numerator by the reciprocal of the denominator.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{2}{2 - \sqrt{3}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.11

Cancel the common factor of $2$ .

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Step 2.3.1.4.11.1

Cancel the common factor.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{2}{2 - \sqrt{3}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.11.2

Rewrite the expression.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{(2 + \sqrt{3}) \frac{1}{2 - \sqrt{3}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.12

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{(2 + \sqrt{3}) (\frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}})}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.13

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{(2 - \sqrt{3}) (2 + \sqrt{3})}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.14

Expand the denominator using the FOIL method.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{4 + 2 \sqrt{3} - 2 \sqrt{3} - {\sqrt{3}}^{2}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.15

Simplify.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{1}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.16

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{(2 + \sqrt{3}) (2 + \sqrt{3})}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.17

Expand $(2 + \sqrt{3}) (2 + \sqrt{3})$ using the FOIL Method.

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Step 2.3.1.4.17.1

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{2 (2 + \sqrt{3}) + \sqrt{3} (2 + \sqrt{3})}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.17.2

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} (2 + \sqrt{3})}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.17.3

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18

Simplify and combine like terms.

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Step 2.3.1.4.18.1

Simplify each term.

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Step 2.3.1.4.18.1.1

Multiply $2$ by $2$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{4 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.1.2

Move $2$ to the left of $\sqrt{3}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{4 + 2 \sqrt{3} + 2 \cdot \sqrt{3} + \sqrt{3} \sqrt{3}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.1.3

Combine using the product rule for radicals.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3 \cdot 3}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.1.4

Multiply $3$ by $3$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{9}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.1.5

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3^{2}}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.1.6

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + 3}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.2

Add $4$ and $3$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 2 \sqrt{3} + 2 \sqrt{3}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.1.4.18.3

Add $2 \sqrt{3}$ and $2 \sqrt{3}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - \tan (\frac{7 π}{12}))}$

Step 2.3.2

The exact value of $\tan (\frac{7 π}{12})$ is $- \sqrt{7 + 4 \sqrt{3}}$ .

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Step 2.3.2.1

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - \tan (\frac{\frac{7 π}{6}}{2}))}$

Step 2.3.2.2

Apply the tangent half-angle identity.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - (\pm \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}}))}$

Step 2.3.2.3

Change the $\pm$ to $-$ because tangent is negative in the second quadrant.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4

Simplify $- \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{1 + \cos (\frac{7 π}{6})}}$ .

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Step 2.3.2.4.1

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{1 - - \cos (\frac{π}{6})}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4.3

Multiply $- - \frac{\sqrt{3}}{2}$ .

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Step 2.3.2.4.3.1

Multiply $- 1$ by $- 1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4.3.2

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4.4

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4.5

Combine the numerators over the common denominator.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{7 π}{6})}})}$

Step 2.3.2.4.6

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \cos (\frac{π}{6})}})}$

Step 2.3.2.4.7

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}})}$

Step 2.3.2.4.8

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2}{2} - \frac{\sqrt{3}}{2}}})}$

Step 2.3.2.4.9

Combine the numerators over the common denominator.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}})}$

Step 2.3.2.4.10

Multiply the numerator by the reciprocal of the denominator.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{2}{2 - \sqrt{3}}})}$

Step 2.3.2.4.11

Cancel the common factor of $2$ .

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Step 2.3.2.4.11.1

Cancel the common factor.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{2}{2 - \sqrt{3}}})}$

Step 2.3.2.4.11.2

Rewrite the expression.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{(2 + \sqrt{3}) \frac{1}{2 - \sqrt{3}}})}$

Step 2.3.2.4.12

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{(2 + \sqrt{3}) (\frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}})})}$

Step 2.3.2.4.13

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{(2 - \sqrt{3}) (2 + \sqrt{3})}})}$

Step 2.3.2.4.14

Expand the denominator using the FOIL method.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{4 + 2 \sqrt{3} - 2 \sqrt{3} - {\sqrt{3}}^{2}}})}$

Step 2.3.2.4.15

Simplify.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{(2 + \sqrt{3}) \frac{2 + \sqrt{3}}{1}})}$

Step 2.3.2.4.16

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{(2 + \sqrt{3}) (2 + \sqrt{3})})}$

Step 2.3.2.4.17

Expand $(2 + \sqrt{3}) (2 + \sqrt{3})$ using the FOIL Method.

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Step 2.3.2.4.17.1

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{2 (2 + \sqrt{3}) + \sqrt{3} (2 + \sqrt{3})})}$

Step 2.3.2.4.17.2

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} (2 + \sqrt{3})})}$

Step 2.3.2.4.17.3

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{2 \cdot 2 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}})}$

Step 2.3.2.4.18

Simplify and combine like terms.

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Step 2.3.2.4.18.1

Simplify each term.

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Step 2.3.2.4.18.1.1

Multiply $2$ by $2$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{4 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}})}$

Step 2.3.2.4.18.1.2

Move $2$ to the left of $\sqrt{3}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{4 + 2 \sqrt{3} + 2 \cdot \sqrt{3} + \sqrt{3} \sqrt{3}})}$

Step 2.3.2.4.18.1.3

Combine using the product rule for radicals.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3 \cdot 3}})}$

Step 2.3.2.4.18.1.4

Multiply $3$ by $3$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{9}})}$

Step 2.3.2.4.18.1.5

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3^{2}}})}$

Step 2.3.2.4.18.1.6

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

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + 3})}$

Step 2.3.2.4.18.2

Add $4$ and $3$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{7 + 2 \sqrt{3} + 2 \sqrt{3}})}$

Step 2.3.2.4.18.3

Add $2 \sqrt{3}$ and $2 \sqrt{3}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 - - \sqrt{7 + 4 \sqrt{3}})}$

Step 2.3.3

Multiply $- - \sqrt{7 + 4 \sqrt{3}}$ .

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Step 2.3.3.1

Multiply $- 1$ by $- 1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 + 1 \sqrt{7 + 4 \sqrt{3}})}$

Step 2.3.3.2

Multiply $\sqrt{7 + 4 \sqrt{3}}$ by $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{(1 - \sqrt{7 + 4 \sqrt{3}}) (1 + \sqrt{7 + 4 \sqrt{3}})}$

Step 3

Expand $(1 - \sqrt{7 + 4 \sqrt{3}}) (1 + \sqrt{7 + 4 \sqrt{3}})$ using the FOIL Method.

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Step 3.1

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 (1 + \sqrt{7 + 4 \sqrt{3}}) - \sqrt{7 + 4 \sqrt{3}} (1 + \sqrt{7 + 4 \sqrt{3}})}$

Step 3.2

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 \cdot 1 + 1 \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} (1 + \sqrt{7 + 4 \sqrt{3}})}$

Step 3.3

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 \cdot 1 + 1 \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} \cdot 1 - \sqrt{7 + 4 \sqrt{3}} \sqrt{7 + 4 \sqrt{3}}}$

Step 4

Simplify and combine like terms.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $1$ by $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + 1 \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} \cdot 1 - \sqrt{7 + 4 \sqrt{3}} \sqrt{7 + 4 \sqrt{3}}}$

Step 4.1.2

Multiply $\sqrt{7 + 4 \sqrt{3}}$ by $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} \cdot 1 - \sqrt{7 + 4 \sqrt{3}} \sqrt{7 + 4 \sqrt{3}}}$

Step 4.1.3

Multiply $- 1$ by $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} \sqrt{7 + 4 \sqrt{3}}}$

Step 4.1.4

Multiply $- \sqrt{7 + 4 \sqrt{3}} \sqrt{7 + 4 \sqrt{3}}$ .

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Step 4.1.4.1

Raise $\sqrt{7 + 4 \sqrt{3}}$ to the power of $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - ({\sqrt{7 + 4 \sqrt{3}}}^{1} \sqrt{7 + 4 \sqrt{3}})}$

Step 4.1.4.2

Raise $\sqrt{7 + 4 \sqrt{3}}$ to the power of $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - ({\sqrt{7 + 4 \sqrt{3}}}^{1} {\sqrt{7 + 4 \sqrt{3}}}^{1})}$

Step 4.1.4.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {\sqrt{7 + 4 \sqrt{3}}}^{1 + 1}}$

Step 4.1.4.4

Add $1$ and $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {\sqrt{7 + 4 \sqrt{3}}}^{2}}$

Step 4.1.5

Rewrite ${\sqrt{7 + 4 \sqrt{3}}}^{2}$ as $7 + 4 \sqrt{3}$ .

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Step 4.1.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7 + 4 \sqrt{3}}$ as ${(7 + 4 \sqrt{3})}^{\frac{1}{2}}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {({(7 + 4 \sqrt{3})}^{\frac{1}{2}})}^{2}}$

Step 4.1.5.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {(7 + 4 \sqrt{3})}^{\frac{1}{2} \cdot 2}}$

Step 4.1.5.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {(7 + 4 \sqrt{3})}^{\frac{2}{2}}}$

Step 4.1.5.4

Cancel the common factor of $2$ .

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Step 4.1.5.4.1

Cancel the common factor.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {(7 + 4 \sqrt{3})}^{\frac{2}{2}}}$

Step 4.1.5.4.2

Rewrite the expression.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - {(7 + 4 \sqrt{3})}^{1}}$

Step 4.1.5.5

Simplify.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - (7 + 4 \sqrt{3})}$

Step 4.1.6

Apply the distributive property.

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - 1 \cdot 7 - (4 \sqrt{3})}$

Step 4.1.7

Multiply $- 1$ by $7$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - 7 - (4 \sqrt{3})}$

Step 4.1.8

Multiply $4$ by $- 1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{1 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - 7 - 4 \sqrt{3}}$

Step 4.2

Subtract $7$ from $1$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{- 6 + \sqrt{7 + 4 \sqrt{3}} - \sqrt{7 + 4 \sqrt{3}} - 4 \sqrt{3}}$

Step 4.3

Subtract $\sqrt{7 + 4 \sqrt{3}}$ from $\sqrt{7 + 4 \sqrt{3}}$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{- 6 + 0 - 4 \sqrt{3}}$

Step 4.4

Add $- 6$ and $0$ .

$\frac{- 2 \sqrt{7 + 4 \sqrt{3}}}{- 6 - 4 \sqrt{3}}$

Step 5

Cancel the common factor of $- 2$ and $- 6 - 4 \sqrt{3}$ .

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Step 5.1

Factor $2$ out of $- 2 \sqrt{7 + 4 \sqrt{3}}$ .

$\frac{2 (- \sqrt{7 + 4 \sqrt{3}})}{- 6 - 4 \sqrt{3}}$

Step 5.2

Cancel the common factors.

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Step 5.2.1

Factor $2$ out of $- 6$ .

$\frac{2 (- \sqrt{7 + 4 \sqrt{3}})}{2 (- 3) - 4 \sqrt{3}}$

Step 5.2.2

Factor $2$ out of $- 4 \sqrt{3}$ .

$\frac{2 (- \sqrt{7 + 4 \sqrt{3}})}{2 (- 3) + 2 (- 2 \sqrt{3})}$

Step 5.2.3

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

$\frac{2 (- \sqrt{7 + 4 \sqrt{3}})}{2 (- 3 - 2 \sqrt{3})}$

Step 5.2.4

Cancel the common factor.

$\frac{2 (- \sqrt{7 + 4 \sqrt{3}})}{2 (- 3 - 2 \sqrt{3})}$

Step 5.2.5

Rewrite the expression.

$\frac{- \sqrt{7 + 4 \sqrt{3}}}{- 3 - 2 \sqrt{3}}$

Step 6

Move the negative in front of the fraction.

$- \frac{\sqrt{7 + 4 \sqrt{3}}}{- 3 - 2 \sqrt{3}}$

Step 7

Multiply $\frac{\sqrt{7 + 4 \sqrt{3}}}{- 3 - 2 \sqrt{3}}$ by $\frac{- 3 + 2 \sqrt{3}}{- 3 + 2 \sqrt{3}}$ .

$- (\frac{\sqrt{7 + 4 \sqrt{3}}}{- 3 - 2 \sqrt{3}} \cdot \frac{- 3 + 2 \sqrt{3}}{- 3 + 2 \sqrt{3}})$

Step 8

Multiply $\frac{\sqrt{7 + 4 \sqrt{3}}}{- 3 - 2 \sqrt{3}}$ by $\frac{- 3 + 2 \sqrt{3}}{- 3 + 2 \sqrt{3}}$ .

$- \frac{\sqrt{7 + 4 \sqrt{3}} (- 3 + 2 \sqrt{3})}{(- 3 - 2 \sqrt{3}) (- 3 + 2 \sqrt{3})}$

Step 9

Expand the denominator using the FOIL method.

$- \frac{\sqrt{7 + 4 \sqrt{3}} (- 3 + 2 \sqrt{3})}{9 - 6 \sqrt{3} + 6 \sqrt{3} - 4 {\sqrt{3}}^{2}}$

Step 10

Simplify.

$- \frac{\sqrt{7 + 4 \sqrt{3}} (- 3 + 2 \sqrt{3})}{- 3}$

Step 11

Apply the distributive property.

$- \frac{\sqrt{7 + 4 \sqrt{3}} \cdot - 3 + \sqrt{7 + 4 \sqrt{3}} (2 \sqrt{3})}{- 3}$

Step 12

Move $- 3$ to the left of $\sqrt{7 + 4 \sqrt{3}}$ .

$- \frac{- 3 \cdot \sqrt{7 + 4 \sqrt{3}} + \sqrt{7 + 4 \sqrt{3}} (2 \sqrt{3})}{- 3}$

Step 13

Combine using the product rule for radicals.

$- \frac{- 3 \sqrt{7 + 4 \sqrt{3}} + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{- 3}$

Step 14

Move the negative in front of the fraction.

$- - \frac{- 3 \sqrt{7 + 4 \sqrt{3}} + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$

Step 15

Multiply $- - \frac{- 3 \sqrt{7 + 4 \sqrt{3}} + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$ .

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Step 15.1

Multiply $- 1$ by $- 1$ .

$1 \frac{- 3 \sqrt{7 + 4 \sqrt{3}} + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$

Step 15.2

Multiply $\frac{- 3 \sqrt{7 + 4 \sqrt{3}} + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$ by $1$ .

$\frac{- 3 \sqrt{7 + 4 \sqrt{3}} + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$

Step 16

Factor $- 1$ out of $- 3 \sqrt{7 + 4 \sqrt{3}}$ .

$\frac{- (3 \sqrt{7 + 4 \sqrt{3}}) + 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$

Step 17

Factor $- 1$ out of $2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}$ .

$\frac{- (3 \sqrt{7 + 4 \sqrt{3}}) - (- 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3})}{3}$

Step 18

Factor $- 1$ out of $- (3 \sqrt{7 + 4 \sqrt{3}}) - (- 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3})$ .

$\frac{- (3 \sqrt{7 + 4 \sqrt{3}} - 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3})}{3}$

Step 19

Simplify the expression.

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Step 19.1

Rewrite $- (3 \sqrt{7 + 4 \sqrt{3}} - 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3})$ as $- 1 (3 \sqrt{7 + 4 \sqrt{3}} - 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3})$ .

$\frac{- 1 (3 \sqrt{7 + 4 \sqrt{3}} - 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3})}{3}$

Step 19.2

Move the negative in front of the fraction.

$- \frac{3 \sqrt{7 + 4 \sqrt{3}} - 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$

Step 20

The result can be shown in multiple forms.

Exact Form:

$- \frac{3 \sqrt{7 + 4 \sqrt{3}} - 2 \sqrt{(7 + 4 \sqrt{3}) \cdot 3}}{3}$

Decimal Form:

$0.57735026 \dots$