Question

The equation whose roots are ${\tan }^2\frac{2\pi }{7},{\tan }^2\frac{4\pi }{7},{\tan }^2\frac{6\pi }{7}$ is

• A. $x^3-21x^2+35x-7=0$
• B. $x^3-35x^2+21x-7=0$
• C. $x^3-35x^2+7x-21=0$
• D. $x^3-21x^2+7x-35=0$

Answer 1

The correct option is A $x^3-21x^2+35x-7=0$

$\cos \frac{2\pi }{7},\cos \frac{4\pi }{7},\cos \frac{6\pi }{7}$ are roots of $8x^3+4x^2-4x-1=0$
Then
$\frac{1}{\cos \frac{2\pi }{7}},\frac{1}{\cos \frac{4\pi }{7}},\frac{1}{\cos \frac{6\pi }{7}}$ are roots of $8{\left(\frac{1}{x}\right)}^3+4{\left(\frac{1}{x}\right)}^2-4\left(\frac{1}{x}\right)-1=0$
$\Rightarrow \sec \frac{2\pi }{7},\sec \frac{4\pi }{7},\sec \frac{6\pi }{7}$ are roots of $x^3+4x^2-4x-8=0$
Now let $y={\tan }^2\theta$ where $\theta =\frac{2\pi }{7},\frac{4\pi }{7},\frac{6\pi }{7}$
$y={\tan }^2\theta ={\sec }^2\theta -1=x^2-1\Rightarrow x^2=1+y$
From
$x^3+4x^2-4x-8=0\Rightarrow x^3-4x=-4x^2+8\Rightarrow x(x^2-4)=-4(x^2-2)\Rightarrow x^2{(x^2-4)}^2=16{(x^2-2)}^2\Rightarrow (1+y){({(1+y)}^2-4)}^2=16{({(1+y)}^2-2)}^2\Rightarrow y^3-21y^2+35y-7=0$
Therefore
${\tan }^2\frac{2\pi }{7},{\tan }^2\frac{4\pi }{7},{\tan }^2\frac{6\pi }{7}$ are roots of $y^3-21y^2+35y-7=0$