1sir can u plz elaborate ur technique
how is
where 
i.e is how is
\prod_{k=1}^{n}{\frac{1}{x^2-2x\cos \alpha+1}}=\frac{1}{n}\sum_{k=1}^{n}{\frac{1-x\cos \alpha}{x^2-2x\cos \alpha+1}}
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5 Answers
62think of how you would break \frac{5x^3+11x+3}{(x^2+1)(x^2+4)}=\frac{2x+1}{x^2+1}+\frac{3x-1}{x^2+4} as partial fraction?
1\frac{5x^3+11x+3}{(x^2+1)(x^2+4)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4} \\ puttin \ x=0 \\ 3=4B+D.........1\\ putting \ x=1 \\ {19}=5(A+B)+2(C+D)\\ putting \ x=-1 \\ -13=5(-A+B)+2(-C+D)\\ 3=5B+2D..........2 \\ solving \ 1\ and \ 2 \ we \ get \\ B=1 ,D=-1
similarly we can find A and C
62umm.. no not this way.. i was trying to break it by substituing the roots!
Like x=i
62\\\frac{5x^3+11x+3}{(x^2+1)(x^2+4)}=\frac{2x+1}{x^2+1}+\frac{3x-1}{x^2+4} \\5x^3+11x+3=(Ax+B)(x^2+4)+(Cx+D)(x^2+1)
now x=i gives -5i+11i+3=(Ai+B)(3)
2i+1=(Ai+B)
A=2, B=1
What this method does is that it eliminates the need for solving simultaneous Equations.