a.maths question.....

tan2x = 2tanx / [(1 - (tanx)^2] = 2y/(1 - y^2)...(1) , (y = tanx)
tan3x = (tan2x + tanx)/(1 - tan2x tanx)
= {2tanx / [(1 - (tanx)^2] + (tanx)} / {1 - 2tanx tanx / [(1 - (tanx)^2]}
= [2y / (1 - y^2) + y] / [1 - (2y^2) / (1 - y^2)]
= [2y + y(1 - y^2)] / [ (1 - y^2) - 2y^2]
= y(3 - y^2) / (1 - 3y^2) , (y = tanx)
RHS = (tan3x - tan2x - tanx) / (tan3x tanx)
=[ y(3 - y^2) / (1 - 3y^2) - 2y/(1 - y^2) - y ]/ [y * y(3 - y^2) / (1 - 3y^2)]
= [(3 - y^2) / (1 - 3y^2) - 2/(1 - y^2) - 1]/ [ y(3 - y^2) / (1 - 3y^2)]
= [(3 - y^2) - 2(1 - 3y^2)/(1 - y^2) - (1 - 3y^2)] / y(3 - y^2)
= [1 - 2(1 - 3y^2)/[(1 - y^2)(3 - y^2)] - (1 - 3y^2)/(3 - y^2)] / y
= [(1 - y^2)(3 - y^2) - 2(1 - 3y^2) - (1 - 3y^2)(1 - y^2)] / [y(1 - y^2)(3 - y^2)]
= (3 - 4y^2 + y^4 - 2 + 6y^2 - 1 + 4y^2 - 3y^4) / [y(1 - y^2)(3 - y^2)]
= (-2y^4 + 6y^2) / [y(1 - y^2)(3 - y^2)]
= (-2y^3 + 6y) / [(1 - y^2)(3 - y^2)]
= 2y( 3 - y^2) / (1 - y^2)(3 - y^2)
= 2y / (1 - y^2)
By (1) = tan 2A = LHS




2010-01-17 15:18:55 補充：
The best method :

By tan 3x

= tan(x + 2x)

= (tan x + tan 2x) / (1 - (tan x)(tan 2x)) ,

we have

(tan3x) ( 1 - (tanx)(tan2x) ) = tan x + tan 2x

tan3x - (tanx)(tan2x)(tan3x) = tan x + tan 2x

tan3x - tan2x - tanx = (tanx)(tan2x)(tan3x)

(tan3x - tan2x - tanx) / [(tanx)(tan3x)] = tan2x

2010-01-17 15:19:15 補充：
The best method :

By tan 3x

= tan(x + 2x)

= (tan x + tan 2x) / (1 - (tan x)(tan 2x)) ,

we have

(tan3x) ( 1 - (tanx)(tan2x) ) = tan x + tan 2x

tan3x - (tanx)(tan2x)(tan3x) = tan x + tan 2x

tan3x - tan2x - tanx = (tanx)(tan2x)(tan3x)

(tan3x - tan2x - tanx) / [(tanx)(tan3x)] = tan2x