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Find tan((7pi)/(12)) exactly using an angle addition or subtraction formula. [Hint: This diagram of special trigonometry. values may help.]. Choose 1 answer: (A) (1-sqrt3)/(1+sqrt3) (B) (-1-sqrt3)/(1-sqrt3) (c) (-1+sqrt3)/(1+sqrt3) (D) (1+sqrt3)/(1-sqrt3)

Find tan⁡(7π12) \tan \left(\frac{7 \pi}{12}\right) exactly using an angle addition or subtraction formula.\newline[Hint: This diagram of special trigonometry. values may help.].\newlineChoose 11 answer:\newline(A) 1−31+3 \frac{1-\sqrt{3}}{1+\sqrt{3}} \newline(B) −1−31−3 \frac{-1-\sqrt{3}}{1-\sqrt{3}} \newline(C) −1+31+3 \frac{-1+\sqrt{3}}{1+\sqrt{3}} \newline(D) 1+31−3 \frac{1+\sqrt{3}}{1-\sqrt{3}}

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Q. Find tan⁡(7π12) \tan \left(\frac{7 \pi}{12}\right) exactly using an angle addition or subtraction formula.\newline[Hint: This diagram of special trigonometry. values may help.].\newlineChoose 11 answer:\newline(A) 1−31+3 \frac{1-\sqrt{3}}{1+\sqrt{3}} \newline(B) −1−31−3 \frac{-1-\sqrt{3}}{1-\sqrt{3}} \newline(C) −1+31+3 \frac{-1+\sqrt{3}}{1+\sqrt{3}} \newline(D) 1+31−3 \frac{1+\sqrt{3}}{1-\sqrt{3}}
  1. Recognize Non-Standard Angle: Recognize that (7π)/12(7\pi)/12 is not a standard angle for which we have a direct trigonometric value. We need to express (7π)/12(7\pi)/12 as the sum or difference of angles for which we do have standard trigonometric values.
  2. Express as Sum of Angles: Notice that (7π)/12(7\pi)/12 can be written as the sum of (4π)/12(4\pi)/12 and (3π)/12(3\pi)/12, which simplifies to π/3+π/4\pi/3 + \pi/4. These are angles for which we have known trigonometric values.
  3. Apply Angle Addition Formula: Use the angle addition formula for tangent, which is tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}, where AA is π/3\pi/3 and BB is π/4\pi/4.
  4. Find Tangent Values: Find the tangent values for AA and BB. We know that tan⁡(π3)=3\tan(\frac{\pi}{3}) = \sqrt{3} and tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1.
  5. Substitute Values: Substitute the values into the angle addition formula: tan⁡(π3+π4)=tan⁡(π3)+tan⁡(π4)1−tan⁡(π3)tan⁡(π4)=3+11−3⋅1.\tan(\frac{\pi}{3} + \frac{\pi}{4}) = \frac{\tan(\frac{\pi}{3}) + \tan(\frac{\pi}{4})}{1 - \tan(\frac{\pi}{3})\tan(\frac{\pi}{4})} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}\cdot 1}.
  6. Simplify Expression: Simplify the expression: (3+1)/(1−3)=(3+1)/(1−3)×((1+3)/(1+3))(\sqrt{3} + 1) / (1 - \sqrt{3}) = (\sqrt{3} + 1) / (1 - \sqrt{3}) \times ((1 + \sqrt{3}) / (1 + \sqrt{3})) to rationalize the denominator.
  7. Rationalize Denominator: Multiply the numerators and the denominators: (3+1)(1+3)/(1−3)(1+3)=(3+3+3+1)/(1−3)(\sqrt{3} + 1)(1 + \sqrt{3}) / (1 - \sqrt{3})(1 + \sqrt{3}) = (3 + \sqrt{3} + \sqrt{3} + 1) / (1 - 3).
  8. Multiply Numerators and Denominators: Simplify the numerator and the denominator: (4+23)/(−2)=2(2+3)/(−2)=−(2+3)(4 + 2\sqrt{3}) / (-2) = 2(2 + \sqrt{3}) / (-2) = - (2 + \sqrt{3}).
  9. Final Simplified Expression: The final simplified expression is −(2+3)- (2 + \sqrt{3}), which corresponds to choice (D) 1+31−3\frac{1 + \sqrt{3}}{1 - \sqrt{3}} if we consider the negative sign as part of the fraction.

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