Find $\tan \left(\frac{19 \pi}{12}\right)$ exactly using an angle addition or subtraction formula. $\newline$ [Hint: This diagram of special trigonometry. values may help.]. $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{3-\sqrt{3}}{3+\sqrt{3}}$ $\newline$ (B) $\frac{3-\sqrt{3}}{3-\sqrt{3}}$ $\newline$ (C) $\frac{-3+\sqrt{3}}{3+\sqrt{3}}$ $\newline$ (D) $\frac{3+\sqrt{3}}{-3+\sqrt{3}}$

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Algebra 1

Domain and range of relations

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Q. Find

\tan \left(\frac{19 \pi}{12}\right)

exactly using an angle addition or subtraction formula.

\newline

[Hint: This diagram of special trigonometry. values may help.].

\newline

Choose

1

answer:

\newline

(A)

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

\newline

(B)

\frac{3-\sqrt{3}}{3-\sqrt{3}}

\newline

(C)

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

\newline

(D)

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

Express as Sum/Difference: We need to express $(19\pi/12)$ as a sum or difference of angles for which we know the exact trigonometric values. The angles that are commonly used and for which we have exact values are $\pi/6$ , $\pi/4$ , and $\pi/3$ (and their multiples). We can write $(19\pi/12)$ as the sum of $(16\pi/12)$ and $(3\pi/12)$ , which simplifies to $(4\pi/3) + (\pi/4)$ .
Apply Angle Addition Formula: Now we will use the angle addition formula for tangent, which is $\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}$ . Let $\alpha = \frac{4\pi}{3}$ and $\beta = \frac{\pi}{4}$ .
Find Tangent Values: We need to find the exact values for $tan(\frac{4\pi}{3})$ and $tan(\frac{\pi}{4})$ . From the unit circle, we know that $tan(\frac{\pi}{4}) = 1$ . For $tan(\frac{4\pi}{3})$ , we know that $\frac{4\pi}{3}$ is in the third quadrant where tangent is positive, and $tan(\frac{4\pi}{3})$ is the same as $tan(\frac{\pi}{3})$ because they differ by an exact circle $(2\pi)$ . Since $tan(\frac{\pi}{3}) = \sqrt{3}$ , $tan(\frac{4\pi}{3}) = -\sqrt{3}$ (because it's in the third quadrant where tangent is negative).
Substitute Values: Now we substitute these values into the angle addition formula: $\tan\left(\frac{$19$\pi}{$12$}\right) = \tan\left(\frac{$4$\pi}{$3$} + \frac{\pi}{$4$}\right) = \frac{\tan\left(\frac{$4$\pi}{$3$}\right) + \tan\left(\frac{\pi}{$4$}\right)}{$1$ - \tan\left(\frac{$4$\pi}{$3$}\right)\tan\left(\frac{\pi}{$4$}\right)} = \frac{(-\sqrt{$3$}) + $1$}{$1$ - (-\sqrt{$3$})($1$)}.
Simplify Expression: Simplify the expression: $\tan\left(\frac{19\pi}{12}\right) = \frac{(-\sqrt{3}) + 1}{1 + \sqrt{3}}$.
Rationalize Denominator: To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 - \sqrt{3})$.
Perform Multiplication: Perform the multiplication: $\left((-\sqrt{3} + 1) \times (1 - \sqrt{3})\right) / \left((1 + \sqrt{3}) \times (1 - \sqrt{3})\right)$.
Use Difference of Squares: Use the difference of squares formula for the denominator: $(1 - \sqrt{3})^2 = 1^2 - 2\cdot 1\cdot \sqrt{3} + (\sqrt{3})^2 = 1 - 2\sqrt{3} + 3$.
Expand Numerator: Simplify the denominator: $1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3}$.
Final Simplification: Expand the numerator: $(-\sqrt{3} + 1) \times (1 - \sqrt{3}) = -\sqrt{3} + \sqrt{3}^2 - 1 + \sqrt{3} = -1 + 3 - \sqrt{3} = 2 - \sqrt{3}.$
Correct Division: Now we have: $\tan\left(\frac{19\pi}{12}\right) = \frac{2 - \sqrt{3}}{4 - 2\sqrt{3}}$.
Correct Division: Now we have: $\tan\left(\frac{19\pi}{12}\right) = \frac{2 - \sqrt{3}}{4 - 2\sqrt{3}}$.To simplify the fraction, we divide both the numerator and the denominator by $2$: $\frac{2 - \sqrt{3}}{2} / \frac{4 - 2\sqrt{3}}{2} = \frac{1 - \sqrt{3}/2}{2 - \sqrt{3}}$.
Correct Division: Now we have: $\tan\left(\frac{19\pi}{12}\right) = \frac{2 - \sqrt{3}}{4 - 2\sqrt{3}}$.To simplify the fraction, we divide both the numerator and the denominator by $2$: $\frac{2 - \sqrt{3}}{2} / \frac{4 - 2\sqrt{3}}{2} = \frac{1 - \sqrt{3}/2}{2 - \sqrt{3}}$.We made a mistake in the previous step by dividing incorrectly. We should divide both the numerator and the denominator by $2$ correctly: $\frac{2 - \sqrt{3}}{2} / \frac{4 - 2\sqrt{3}}{2} = \frac{1 - \sqrt{3}/2}{2 - \sqrt{3}}$.