Question

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

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

Answer 1

The correct option is A $-4$

As $\cos \frac{2\pi }{7},\cos \frac{4\pi }{7},\cos \frac{6\pi }{7}$ roots of $8x^3+4x^2-4x-1=0$
Then sum of roots
$\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=-4$
And product of roots
$\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{6\pi }{7}=1$
Therefore
$\sec \frac{2\pi }{7}\sec \frac{4\pi }{7}+\sec \frac{4\pi }{7}\sec \frac{6\pi }{7}+\sec \frac{6\pi }{7}\sec \frac{2\pi }{7}=\frac{1}{\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}}+\frac{1}{\cos \frac{4\pi }{7}\cos \frac{6\pi }{7}}+\frac{1}{\cos \frac{6\pi }{7}\cos \frac{2\pi }{7}}=\frac{\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}}{\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{6\pi }{7}}=\frac{-4}{1}=-4$