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What is the exact value of tan⁡(19π12)\tan(\frac{19\pi}{12})?

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Q. What is the exact value of tan⁡(19π12)\tan(\frac{19\pi}{12})?
  1. Break down angle: Break down the angle 19π12\frac{19\pi}{12} into a sum or difference of angles that are easier to work with, such as π\pi and π4\frac{\pi}{4} or π3\frac{\pi}{3} and π6\frac{\pi}{6}, since the tangent values for π4\frac{\pi}{4}, π3\frac{\pi}{3}, and π6\frac{\pi}{6} are known.\newline19π12\frac{19\pi}{12} can be expressed as 16π12+3π12\frac{16\pi}{12} + \frac{3\pi}{12}, which simplifies to π\pi00.
  2. Use angle sum identity: Use the angle sum identity for tangent, which states that tan⁡(α+β)=tan⁡(α)+tan⁡(β)1−tan⁡(α)tan⁡(β)\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}, where α=4π3\alpha = \frac{4\pi}{3} and β=π4\beta = \frac{\pi}{4}.
  3. Find tangent values: Find the exact values for tan⁡(4π3)\tan(\frac{4\pi}{3}) and tan⁡(π4)\tan(\frac{\pi}{4}).\newlinetan⁡(4π3)\tan(\frac{4\pi}{3}) is the tangent of an angle in the third quadrant, where tangent is positive. The reference angle for 4π3\frac{4\pi}{3} is π3\frac{\pi}{3}, and tan⁡(π3)=3\tan(\frac{\pi}{3}) = \sqrt{3}. Therefore, tan⁡(4π3)=3\tan(\frac{4\pi}{3}) = \sqrt{3}.\newlinetan⁡(π4)\tan(\frac{\pi}{4}) is the tangent of an angle in the first quadrant, where tangent is positive, and tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1.
  4. Substitute values: Substitute the values into the angle sum identity.\newlinetan⁡(19π12)=tan⁡(4π3+π4)=(tan⁡(4π3)+tan⁡(π4))(1−tan⁡(4π3)tan⁡(π4))\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}))}\newlinetan⁡(19π12)=(3+1)(1−3⋅1)\tan(\frac{19\pi}{12}) = \frac{(\sqrt{3} + 1)}{(1 - \sqrt{3} \cdot 1)}
  5. Simplify expression: Simplify the expression.\newlinetan⁡(19π12)=3+11−3\tan\left(\frac{19\pi}{12}\right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}\newlineTo rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is (1+3)(1 + \sqrt{3}).
  6. Rationalize denominator: Perform the multiplication to rationalize the denominator.\newlinetan⁡(19π12)=(3+1)×(1+3)/((1−3)×(1+3))\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / ((1 - \sqrt{3}) \times (1 + \sqrt{3}))\newlinetan⁡(19π12)=(3+1)×(1+3)/(1−(3)2)\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / (1 - (\sqrt{3})^2)\newlinetan⁡(19π12)=(3+1)×(1+3)/(1−3)\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / (1 - 3)\newlinetan⁡(19π12)=(3+1)×(1+3)/(−2)\tan\left(\frac{19\pi}{12}\right) = (\sqrt{3} + 1) \times (1 + \sqrt{3}) / (-2)
  7. Expand and simplify: Expand the numerator and simplify the expression.\newlinetan⁡(19π12)=(3⋅1+3⋅3+1⋅1+1⋅3)/(−2)\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)\newlinetan⁡(19π12)=(3+3+1+3)/(−2)\tan\left(\frac{19\pi}{12}\right) = \left(\sqrt{3} + 3 + 1 + \sqrt{3}\right) / (-2)\newlinetan⁡(19π12)=(23+4)/(−2)\tan\left(\frac{19\pi}{12}\right) = \left(2\sqrt{3} + 4\right) / (-2)\newlinetan⁡(19π12)=−3−2\tan\left(\frac{19\pi}{12}\right) = -\sqrt{3} - 2

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