Question

If $7\left(\theta \right)=(2n+1)\pi$, when $n=0,1,2,3,4,5,6,$ then on the basis of given information, answer the given question.

The value of ${\sec }^2\frac{\pi }{7}+{\sec }^2\frac{3\pi }{7}+{\sec }^2\frac{5\pi }{7}$ is,

• A. $24$
• B. $-24$
• C. $80$
• D. $-80$

Answer 1

Answer: (A)

$24$

We know that the roots of $x^4+1=0$ are of the form $\cos \left(\frac{(2n-1)\pi }{7}\right)$ where $n=1,2,3,4$

Also $x^4+1=(x+1)(8x^3-4x^2-4x+1)=0$ $......\left(i\right)$

$\Rightarrow \cos \left(\frac{\pi }{7}\right),\cos \left(\frac{3\pi }{7}\right),\cos \left(\frac{5\pi }{7}\right)$ are the roots of $8x^3-4x^2-4x+1=0$ $...........\left(ii\right)$

Again putting $y=\frac{1}{x^2}$ in $\left(ii\right)$ i.e., $(y={\sec }^2\theta )$

then

$\frac{8}{y^{\frac{3}{2}}}-\frac{4}{y}-\frac{4}{y^{\frac{1}{2}}}+1=0\Rightarrow 8-4y^{\frac{1}{2}}-4y+y^{\frac{3}{2}}=0$

$\Rightarrow y^{1/2}(y-4{)}^2=16(y-2{)}^2\Rightarrow y^3-24y^2+80y-64=0$

has roots ${\sec }^2\frac{\pi }{7}+{\sec }^2\frac{3\pi }{7}+{\sec }^2\frac{5\pi }{7}=24......\left(iv\right)$