Multiple choice

Let $\cos\dfrac {2\pi}{7},\cos\dfrac {4\pi}{7},\cos\dfrac {6\pi}{7}$ be the root of the equation $8x^ {3}+4x^ {2}-4x-1=0$ then the value of $\sec\dfrac {2\pi}{7}\sec\dfrac {4\pi}{7}+\sec\dfrac {2\pi}{7}\sec\dfrac {6\pi}{7}+\sec\dfrac {6\pi}{7}+\sec\dfrac {2\pi}{7}$ is

  1. $-4$
  2. $8$
  3. $\dfrac {-1}{2}$
  4. $\dfrac {-1}{8}$
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A Correct answer
AI explanation

Using Vieta's formulas for the equation 8x^3 + 4x^2 - 4x - 1 = 0, the sum of the roots is -1/2, the sum of the pairwise products is -1/2, and the product of the roots is 1/8. Factoring the intended expression gives sec(2pi/7)sec(4pi/7)sec(6pi/7) times the sum of cos(6pi/7), cos(4pi/7), and cos(2pi/7). This simplifies to (1/cos(a)cos(b)cos(c)) multiplied by (cos(a) + cos(b) + cos(c)), which equals 8 times -1/2. The result is -4.

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