Question

Find the sum : $\tan x\tan 2x+\tan 2x\tan 3x+.....+\tan nx\tan (n+1)x$.

Answer 1

$\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}1+\tan a\tan b=\frac{\tan a-\tan b}{\tan (a-b)}\tan a\tan b=-1+\frac{\tan a-\tan b}{\tan (a-b)}\tan a\tan b=\frac{-\tan (a-b)+\tan a-\tan b}{\tan (a-b)}$

applying this formula to each term we get

$(\tan 2x-\tan x+\tan x+\tan 3x-\tan 2x-\tan x+\tan 4x-\tan 3x-\tan x.....\tan (n+1)x-\tan nx-\tan x)/\tan x\frac{\tan (n+1)x-(n+1)\tan x}{\tan x}$