Question

The value of $\sec \frac{2\pi }{7}+\sec \frac{4\pi }{7}+\sec \frac{6\pi }{7}$ is

• A. $\frac{-1}{2}$
• B. $-4$
• C. $\frac{1}{8}$
• D. $\frac{-1}{8}$

Answer 1

Answer: (B)

$-4$

$\cos x=\frac{1}{\sec \left(x\right)}$
Hence
$\sec \left(x\right)=\frac{1}{\cos \left(x\right)}$
Hence for the required equation, we replace $x$ by $\frac{1}{x}$.
Hence we get
$8(\frac{1}{x}{)}^3+4(\frac{1}{x}{)}^2-4\frac{1}{x}-1=0$
$8+4x-4x^2-x^3=0$
$x^3+4x^2-4x-8=0$
Hence
$\sec \frac{2\pi }{7}+\sec \frac{4\pi }{7}+\sec \frac{6\pi }{7}$
$=$ sum of roots of the above equation
$=-\frac{coefficientof{x}^2}{coefficientof{x}^3}$
$=-4$