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

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Find derivatives of other trigonometric functions

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Q. What is the exact value of

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

?

Express as Sum/Difference: To find the exact value of $\tan\left(\frac{19\pi}{12}\right)$ , we can express $\frac{19\pi}{12}$ as a sum or difference of angles for which we know the exact values of the trigonometric functions. The angles we commonly use are multiples of $\frac{\pi}{6}$ , $\frac{\pi}{4}$ , and $\frac{\pi}{3}$ because their trigonometric values are well known and can be easily calculated.
Apply Angle Sum Identity: We can express $(19\pi)/12$ as the sum of $(16\pi)/12$ and $(3\pi)/12$ , which simplifies to $(4\pi)/3 + \pi/4$ . These two angles are more familiar, and we can use the angle sum identity for tangent to find the exact value of $\tan((19\pi)/12)$ .
Calculate Exact Values: The angle sum identity for tangent is $\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$ . We will use $A = \frac{4\pi}{3}$ and $B = \frac{\pi}{4}$ for our calculation.
Plug into Identity: First, we find the exact values of $\tan(\frac{4\pi}{3})$ and $\tan(\frac{\pi}{4})$ . $\tan(\frac{4\pi}{3})$ is equivalent to $\tan(\pi - \frac{\pi}{3})$ , which is the same as $-\tan(\frac{\pi}{3})$ because tangent is negative in the third quadrant. The exact value of $\tan(\frac{\pi}{3})$ is $\sqrt{3}$ , so $\tan(\frac{4\pi}{3}) = -\sqrt{3}$ . $\tan(\frac{\pi}{4})$ is $1$ because it's the tangent of a $\tan(\frac{\pi}{4})$ $0$ -degree angle.
Simplify Expression: Now we can plug these values into the angle sum identity: $\tan\left(\frac{19\pi}{12}\right) = \tan\left(\frac{4\pi}{3} + \frac{\pi}{4}\right) = \frac{-\sqrt{3} + 1}{1 - (-\sqrt{3} \cdot 1)}$ .
Rationalize Denominator: Simplifying the expression, we get $\tan\left(\frac{19\pi}{12}\right) = \frac{-\sqrt{3} + 1}{1 + \sqrt{3}}$ . To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 - \sqrt{3})$ .
Expand Numerator: Multiplying the numerator and denominator by the conjugate, we get: $\tan\left(\frac{19\pi}{12}\right) = \frac{[-\sqrt{3} + 1][1 - \sqrt{3}]}{[1 + \sqrt{3}][1 - \sqrt{3}]}$ .
Expand Denominator: Expanding the numerator, we get: $(-\sqrt{3} + 1)(1 - \sqrt{3}) = -3 - \sqrt{3} + \sqrt{3} - 1$ . The $\sqrt{3}$ terms cancel out, leaving us with $-3 - 1$ .
Final Simplification: Expanding the denominator, we get: $(1 + \sqrt{3})(1 - \sqrt{3}) = 1 - 3$ , which simplifies to $-2$ .
Final Simplification: Expanding the denominator, we get: $(1 + \sqrt{3})(1 - \sqrt{3}) = 1 - 3$ , which simplifies to $-2$ .Now we have $\tan\left(\frac{19\pi}{12}\right) = \frac{-4}{-2}$ , which simplifies to $2$ .

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