Question

If the roots of the equation are
${\cot }^2\frac{\pi }{2n+1},{\cot }^2\frac{2\pi }{2n+1},{\cot }^2\frac{3\pi }{2n+1},\cdots ,{\cot }^2\frac{n\pi }{2n+1}$,
then show that the value of
${\cot }^2\frac{\pi }{2n+1}+{\cot }^2\frac{2\pi }{2n+1}+{\cot }^2\frac{3\pi }{2n+1}$ $+\cdots +{\cot }^2\frac{n}{2n+1}$ is $\frac{n(2n-1)}{3}.$

Answer 1

In (2) or part (a), replacing n by 2n+1, we get
$\sin (2n+1)\alpha {=}^{2n+1}{C}_1{\cos }^{2n}\alpha \sin {\alpha }^{-2n+1}{C}_3{\cos }^{2n-2}\alpha {\sin }^3\alpha$
${+}^{2n+1}{C}_5{\cos }^{2n-4}\alpha {\sin }^5\alpha -.....={\sin }^{2n+1}\alpha$
$[{}^{2m+1}{C}_1{\cot }^{2n}\alpha {-}^{2n+1}{C}_3{\cot }^{2n-2}\alpha {+}^{2n+1}{C}_5{\cot }^{2n-4}\alpha -\cdots ]\cdots \left(3\right)$
Now L.H.S.=$\sin (2n+1)\alpha =0$
when $(2n+1)=r\pi$
or $\alpha =\frac{\pi }{2n+1},\frac{2\pi }{2n+1},\frac{3\pi }{2n+1},\cdots$
Hence R.H.S. of (3) gives
${}^{2n+1}{C}_1{\cot }^{2n}\alpha {-}^{2n+1}{C}_3{\cot }^{2n-2}\alpha {+}^{2n-4}\alpha -.....=0$
Above in as nth degree equation in ${\cot }^2\alpha .$
Hence ${\cot }^2\frac{\pi }{2n+1},{\cot }^2\frac{2\pi }{2n+1},{\cot }^2\frac{3\pi }{2n+1},...$
${\cot }^2\frac{n\pi }{2n+1}$ are the n roots of the equation
${}^{2n+1}{c}_1{x}^n{-}^{2n+1}{c}_3{x}^{n-1}{+}^{2n+1}{c}_5{x}^{n-2}-....=0$ ...(4)
Sum of the roots of (4)
$=-\frac{coeff.of{x}^{n-1}}{coeff.of{x}^n}=\frac{{}^{2n+1}{c}_3}{{}^{2n+1}{c}_1}$
$=\frac{(2n+1).2n(2n-1)}{\left(\mathrm{1.2.3}\right)(2n+1)}=\frac{n(2n-1)}{3}.$