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What is the exact value of $\tan(\frac{19\pi}{12})$ ?

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Find trigonometric ratios using a Pythagorean or reciprocal identity

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Q. What is the exact value of

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

?

Break down angle: Break down the angle $\frac{19\pi}{12}$ into a sum or difference of angles that are easier to work with, such as $\pi$ and $\frac{\pi}{4}$ or $\frac{\pi}{3}$ and $\frac{\pi}{6}$ , since the tangent values for $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , and $\frac{\pi}{6}$ are known. $\newline$ $\frac{19\pi}{12}$ can be expressed as $\frac{16\pi}{12} + \frac{3\pi}{12}$ , which simplifies to $\pi$ $0$ .
Use angle sum identity: Use the angle sum identity for tangent, which states that $\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}$ , where $\alpha = \frac{4\pi}{3}$ and $\beta = \frac{\pi}{4}$ .
Find tangent values: Find the exact values for $\tan(\frac{4\pi}{3})$ and $\tan(\frac{\pi}{4})$ . $\newline$ $\tan(\frac{4\pi}{3})$ is the tangent of an angle in the third quadrant, where tangent is positive. The reference angle for $\frac{4\pi}{3}$ is $\frac{\pi}{3}$ , and $\tan(\frac{\pi}{3}) = \sqrt{3}$ . Therefore, $\tan(\frac{4\pi}{3}) = \sqrt{3}$ . $\newline$ $\tan(\frac{\pi}{4})$ is the tangent of an angle in the first quadrant, where tangent is positive, and $\tan(\frac{\pi}{4}) = 1$ .
Substitute values: Substitute the values into the angle sum identity. $\newline$ $\tan(\frac{19\pi}{12}) = \tan(\frac{4\pi}{3} + \frac{\pi}{4}) = \frac{(\tan(\frac{4\pi}{3}) + \tan(\frac{\pi}{4}))}{(1 - \tan(\frac{4\pi}{3})\tan(\frac{\pi}{4}))}$ $\newline$ $\tan(\frac{19\pi}{12}) = \frac{(\sqrt{3} + 1)}{(1 - \sqrt{3} \cdot 1)}$
Simplify expression: Simplify the expression. $\newline$ $\tan\left(\frac{19\pi}{12}\right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$ $\newline$ To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 + \sqrt{3})$ .
Rationalize denominator: Perform the multiplication to rationalize the denominator. $\newline$ $\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / ((1 - \sqrt{3}) \times (1 + \sqrt{3}))$ $\newline$ $\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / (1 - (\sqrt{3})^2)$ $\newline$ $\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / (1 - 3)$ $\newline$ $\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / (-2)$
Expand and simplify: Expand the numerator and simplify the expression. $\newline$ $\tan\left(\frac{19\pi}{12}\right) = \left(\sqrt{3} \cdot 1 + \sqrt{3} \cdot \sqrt{3} + 1 \cdot 1 + 1 \cdot \sqrt{3}\right) / (-2)$ $\newline$ $\tan\left(\frac{19\pi}{12}\right) = \left(\sqrt{3} + 3 + 1 + \sqrt{3}\right) / (-2)$ $\newline$ $\tan\left(\frac{19\pi}{12}\right) = \left(2\sqrt{3} + 4\right) / (-2)$ $\newline$ $\tan\left(\frac{19\pi}{12}\right) = -\sqrt{3} - 2$

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