# tan x tan 2x+ tan 2x tan 3x+........................+tan nx tan(n+1)x

68 Points
13 years ago

Dear Divya ,

use the formulae  : tan(a-b) = tan a -tanb /1+tana tanb , we find , tan a tan b  = tana -tan b -tan(a-b) /tan (a-b)

apply to each term we get , expression

= (tan2x -tanx -tanx + tan3x- tan2x -tan x+ tan4x -tan3x -tanx ......tan(n+1)x -tannx -tanx )/tanx

= tan(n+1)x -(n+1)tan x /tanx

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Harsha Kumari
14 Points
5 years ago
tanx + tan2x+ tan3x=0
or, tanx+ tan2x= - tan3x
or, (sinx/cosx)+(sin 2x /Cos 2x)=-(sin 3x /cos 3x)
or, (sin x *cos 2 X + cos x *sin 2x)/cos x * cos 2x = -sin 3x / cos 3x
or, sin( 2x+x)* cos 3x= - cos x *cos 2x *sin 3x
or, sin 3x*cos 3x+ cos x * cos 2x *sin 3x =0
or, sin 3x ( cos 3x + cos x *cos 2x)=0
or, sin 3x (cos ( 2x + x)+ cos x * cos 2x)=0
or, sin 3x (cos x *cos 2x - sin x *sin 2x+ cos x *cos 2x) =0
or, -sin 3x *sin x * sin 2x=0

Either sin3x=0
i.e, 3x=nπ
i.e, X=nπ/3

or, sin 2x=0
i.e, 2x=nπ
i.e, X= nπ/2

or, sin x= 0
i.e, X=nπ
Required solution is X= nπ/3