Trigonometric sum

is it 36 ??

i am not good in trigonometric manupulation , so applying polynomials to it
sec^2(\frac{\pi}{9})+sec^2(\frac{2\pi}{9})+sec^2(\frac{4\pi}{9}) =\left( sec^2(\frac{\pi}{9})+sec^2(\frac{3\pi}{9})+sec^2(\frac{5\pi}{9})+sec^2(\frac{7pi}{9})-4 \right)

\text{Consider the equation}\\ z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1=0 \Leftrightarrow \left(z^4+\frac{1}{z^4} \right)+\left(z^3+\frac{1}{z^3} \right)+\left(z^2+\frac{1}{z^2} \right)+\left(z+\frac{1}{z} \right)+1=0 \\ \text{putting } z+\frac{1}{z}=\frac{2}{x}\ ,\text{we get the following polynomial}\\ x^4+4x^3-12x^2-8x+16=0 \\ \sum{\sec^2\frac{k\pi}{9}}=16-(2*-12)=40
40-4=36

there must be some elegant solution

$Here We have To Calculate value of $sec^2(\frac{\pi}{9})+sec^2(\frac{2\pi}{9})+sec^2(\frac{4\pi}{9})$\\\\ i.e $1+tan^2(\frac{\pi}{9})+1+tan^2(\frac{2\pi}{9})+1+tan^2(\frac{4\pi}{9})$\\\\ $=3+\underbrace{tan^2(\frac{\pi}{9})+tan^2(\frac{2\pi}{9})+tan^2(\frac{4\pi}{9})}$\\\\ Now Here We will Calculate $tan^2(\frac{\pi}{9})+tan^2(\frac{2\pi}{9})+tan^2(\frac{4\pi}{9}).$\\\\ Let $x=tan(\frac{\pi}{9})$. Then $tan(3.\frac{\pi}{9})=\frac{3\tan(\frac{\pi}{9})-tan^3(\frac{\pi}{9})}{1-3tan^2(\frac{\pi}{9})}$\\\\\ $\sqrt{3}=\frac{3x-x^3}{1-3x^2}................(1)$\\\\ $\sqrt{3}=x.(\frac{\sqrt{3}-x}{1+x\sqrt{3}}).(\frac{\sqrt{3}+x}{1-x\sqrt{3}})$\\\\ $\sqrt{3}=\frac{3x-x^3}{1-3x^2}\Leftrightarrow \sqrt{3}-3\sqrt{3}x^2=3x-x^3$\\\\ So $x^3-3\sqrt{3}x^2-3x+\sqrt{3}=0............(2)$ and $\boxed{x^3=3\sqrt{3}x^2+3x-\sqrt{3}}$.\\\\ Multiply equation..(1) by $x$, we Get\\\\
x^4-3\sqrt{3}x^3-3x^2+\sqrt{3}x=0$ and put value of $x^3$ from box, We Get\\\\ $\boxed{x^4-30x^2-8\sqrt{3}x+9=0}...........................(3)$\\\\ and Now from equation..(1), Here We assume $a=tan(\frac{\pi}{9})=x$\\\\ $b=\tan(\frac{2\pi}{9})=tan(\frac{\pi}{3}-\frac{\pi}{9})=(\frac{\sqrt{3}-x}{1+x\sqrt{3}})$\\\\ and $c=\tan(\frac{4\pi}{9})=tan(\frac{\pi}{3}+\frac{\pi}{9})=(\frac{\sqrt{3}+x}{1-x\sqrt{3}})$\\\\ So Put These value in equation..(1), We Get $\sqrt{3}=abc\Leftrightarrow \boxed{bc=\frac{\sqrt{3}}{a}}$\\\\ Now Here We have to calculate $a^2+b^2+c^2......................(4)$\\\\ Here $b=\frac{\sqrt{3}-x}{1+x\sqrt{3}}$ OR $x=\frac{\sqrt{3}-b}{1+b\sqrt{3}}$\\\\ Similarly $c=\frac{\sqrt{3}+x}{1-x\sqrt{3}}$ OR $x=\frac{c-\sqrt{3}}{1+c\sqrt{3}}$\\\\ So $x=\frac{\sqrt{3}-b}{1+b\sqrt{3}}=\frac{c-\sqrt{3}}{1+c\sqrt{3}}$\\\\ $(b+c)=\sqrt{3}(bc-1)\Leftrightarrow \boxed{b^2+c^2=3(bc)^2-8(bc)+3}$\\\\ Put This value in Equation..(4), We Get\\\\
a^2+b^2+c^2=a^2+3(bc)^2-8(bc)+3=a^2+3(\frac{\sqrt{3}}{a})^2-8(\frac{\sqrt{3}}{a})+3$\\\\ $=\frac{a^4+3a^2-8\sqrt{3}a+9}{a^2}=\frac{\underbrace{a^4-8\sqrt{3}a+9}+3a^2}{a^2}=\frac{30a^2+3a^2}{a^2}=33$\\\\ (from equation $x^4-30x^2-8\sqrt{3}x+9=0.$ bcz $x=a$ is a Root of the equation. \\\\ So $a^4-8\sqrt{3}a+9=30a^2)$\\\\ So $a^2+b^2+c^2=33$\\\\ i.e $tan^2(\frac{\pi}{9})+tan^2(\frac{2\pi}{9})+tan^2(\frac{4\pi}{9})=30$\\\\ So $\boxed{\boxed{sec^2(\frac{\pi}{9})+sec^2(\frac{2\pi}{9})+sec^2(\frac{4\pi}{9})=3+tan^2(\frac{\pi}{9})+tan^2(\frac{2\pi}{9})+tan^2(\frac{4\pi}{9})=36}}$\\\\ (Very Lengthy Method......................)

Thanks hsbhatt Sir.

Yes that was the solution i want.......

I am also triying this Method. but have a deep trouble in expanding

tan(nx) = n.tanx - (n,3) tan^3x+....................................

bhatt Sir can u explain me how can i expand function like tan(nx) , sin(nx) , cos(nx)...............

Its there in any standard book like Loney.

The starting point in some books, is by using De Moivre's theorem to obtain

\tan n \theta = \frac{\sin n \theta}{\cos n \theta} = \frac{ (\cos \theta + i \sin \theta)^n - (\cos \theta - i \sin \theta)^n}{(\cos \theta + i \sin \theta)^n + (\cos \theta - i \sin \theta)^n}

prophet sir why are you doing partiality

you did you not answer here http://targetiit.com/iit-jee-forum/posts/permutation-a-neat-problem-18307.html

sorry if i hurt you i didnt mean it that way i hope u get what i m trying to say

for the simple reason that it had not solved the problem myself. ricky had promised a solution and i thought pandit would give the reasoning by which he arrived at that generating function.

As for the derivation of tan nθ, Loney is a widely available book, so i thought i would save some effort. If you guys still need it, I can give it.

Thanks hsbhatt Sir.