Quay lại đáp án

Ta có: $$\tan (n + 1)x - \tan nx = \frac{{\sin (n + 1)x}}{{\cos (n + 1)x}} - \frac{{\sin nx}}{{\cos nx}} = \frac{{\sin (n + 1)x.\cos nx - \sin nx.\cos (n + 1)x}}{{\cos (n + 1)x.\cos nx}} = \frac{{\sin {\rm{[}}(n + 1)x - nx]}}{{\cos (n + 1)x.\cos nx}} = \frac{{\sin x}}{{\cos (n + 1)x.\cos nx}}$$ Suy ra: $\frac{1}{{\cos (n + 1)x.\cos nx}} = \frac{1}{{\sin x}}{\rm{[}}\tan (n + 1)x - \tan nx{\rm{]}}$Từ đó:$A = \frac{1}{{\sin x}}{\rm{[}}\tan 2x - \tan x{\rm{] + }}\frac{1}{{\sin x}}{\rm{[}}\tan 3x - \tan 2x{\rm{] + }}...{\rm{ + }}\frac{1}{{\sin x}}{\rm{[}}\tan 2013x - \tan 2012x{\rm{]}}= \frac{1}{{\sin x}}{\rm{[}}\tan 2013x - \tan x{\rm{]}}$Mà$\tan \frac{{2013\pi }}{{2012}} = \tan (\pi + \frac{\pi }{{2012}}) = \tan \frac{\pi }{{2012}}$Thế$x = \frac{\pi }{{2012}}$vào A ta được:$A = \frac{1}{{\sin \frac{\pi }{{2012}}}}[\tan \frac{{2013\pi }}{{2012}} - \tan \frac{\pi }{{2012}}] = 0$

Ta có: $$\tan (n + 1)x - \tan nx = \frac{{\sin (n + 1)x}}{{\cos (n + 1)x}} - \frac{{\sin nx}}{{\cos nx}} = \frac{{\sin (n + 1)x.\cos nx - \sin nx.\cos (n + 1)x}}{{\cos (n + 1)x.\cos nx}} = \frac{{\sin {\rm{[}}(n + 1)x - nx]}}{{\cos (n + 1)x.\cos nx}} = \frac{{\sin x}}{{\cos (n + 1)x.\cos nx}}$$ Suy ra: $\frac{1}{{\cos nx.\cos (n + 1)x}} = \frac{1}{{\sin x}}[\tan (n + 1)x - \tan nx]$Từ đó:$A = \frac{1}{{\sin x}}{\rm{[}}\tan 2x - \tan x{\rm{] + }}\frac{1}{{\sin x}}{\rm{[}}\tan 3x - \tan 2x{\rm{] + }}...{\rm{ + }}\frac{1}{{\sin x}}{\rm{[}}\tan 2013x - \tan 2012x{\rm{]}}= \frac{1}{{\sin x}}{\rm{[}}\tan 2013x - \tan x{\rm{]}}$Mà$\tan \frac{{2013\pi }}{{2012}} = \tan (\pi + \frac{\pi }{{2012}}) = \tan \frac{\pi }{{2012}}$Thế$x = \frac{\pi }{{2012}}$vào A ta được:$A = \frac{1}{{\sin \frac{\pi }{{2012}}}}[\tan \frac{{2013\pi }}{{2012}} - \tan \frac{\pi }{{2012}}] = 0$