Question

find the sum : $tanxtan2x+tan2xtan3x+.............+tannxtan(n+1)x$.

Answer 1

formula:

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

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

apply the same to each term, we get,

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

$=\frac{\tan (n+1)x-(n+1)\tan x}{\tan x}$