Find the Exact Value tan((7pi)/6+pi/4)

$\tan (\frac{7 π}{6} + \frac{π}{4})$

To write $\frac{7 π}{6}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\tan (\frac{7 π}{6} \cdot \frac{2}{2} + \frac{π}{4})$

To write $\frac{π}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\tan (\frac{7 π}{6} \cdot \frac{2}{2} + \frac{π}{4} \cdot \frac{3}{3})$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{7 π}{6}$ by $\frac{2}{2}$ .

$\tan (\frac{7 π \cdot 2}{6 \cdot 2} + \frac{π}{4} \cdot \frac{3}{3})$

Multiply $6$ by $2$ .

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

Multiply $\frac{π}{4}$ by $\frac{3}{3}$ .

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

Multiply $4$ by $3$ .

$\tan (\frac{7 π \cdot 2}{12} + \frac{π \cdot 3}{12})$

Combine the numerators over the common denominator.

$\tan (\frac{7 π \cdot 2 + π \cdot 3}{12})$

Simplify the numerator.

Multiply $2$ by $7$ .

$\tan (\frac{14 π + π \cdot 3}{12})$

Move $3$ to the left of $π$ .

$\tan (\frac{14 π + 3 \cdot π}{12})$

Add $14 π$ and $3 π$ .

$\tan (\frac{17 π}{12})$

The exact value of $\tan (\frac{17 π}{12})$ is $\sqrt{7 + 4 \sqrt{3}}$ .

Rewrite $\frac{17 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\tan (\frac{\frac{17 π}{6}}{2})$

Apply the tangent half-angle identity.

$\pm \sqrt{\frac{1 - \cos (\frac{17 π}{6})}{1 + \cos (\frac{17 π}{6})}}$

Change the $\pm$ to $+$ because tangent is positive in the third quadrant.

$\sqrt{\frac{1 - \cos (\frac{17 π}{6})}{1 + \cos (\frac{17 π}{6})}}$

Simplify $\sqrt{\frac{1 - \cos (\frac{17 π}{6})}{1 + \cos (\frac{17 π}{6})}}$ .

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

$\sqrt{\frac{1 - \cos (\frac{5 π}{6})}{1 + \cos (\frac{17 π}{6})}}$

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 second quadrant.

$\sqrt{\frac{1 - - \cos (\frac{π}{6})}{1 + \cos (\frac{17 π}{6})}}$

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

$\sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

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

Multiply $- 1$ by $- 1$ .

$\sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

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

$\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

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

$\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{17 π}{6})}}$

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

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{1 + \cos (\frac{5 π}{6})}}$

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

Expand the denominator using the FOIL method.

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

Simplify.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$\sqrt{4 + 2 \sqrt{3} + \sqrt{3} \cdot 2 + \sqrt{3} \sqrt{3}}$

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

$\sqrt{4 + 2 \sqrt{3} + 2 \cdot \sqrt{3} + \sqrt{3} \sqrt{3}}$

Combine using the product rule for radicals.

$\sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3 \cdot 3}}$

Multiply $3$ by $3$ .

$\sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{9}}$

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

$\sqrt{4 + 2 \sqrt{3} + 2 \sqrt{3} + \sqrt{3^{2}}}$

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

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

Add $4$ and $3$ .

$\sqrt{7 + 2 \sqrt{3} + 2 \sqrt{3}}$

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

$\sqrt{7 + 4 \sqrt{3}}$

The result can be shown in multiple forms.

Exact Form:

$\sqrt{7 + 4 \sqrt{3}}$

Decimal Form:

$3.73205080 \dots$

Find the Exact Value tan((7pi)/6+pi/4)

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