can you please help me prove this
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tan(3x)
= tan(x + 2x)
= [ tan x + tan(2x) ] / [ 1 - tan(x) tan(2x)]
= [ tan x + { 2 tan x / (1 - tan^2(x) } ] /[ 1 - tan x { 2 tan x / (1 - tan^2(x) ]
= [tan x (1 - tan^2(x)) + 2tan x ] / [1 - tan^2(x) - 2tan^2(x) ]
= [ 3tan x - tan^3(x) ] / [1 - 3tan^2(x) ]
= tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ]
= tan(x + 2x)
= [ tan x + tan(2x) ] / [ 1 - tan(x) tan(2x)]
= [ tan x + { 2 tan x / (1 - tan^2(x) } ] /[ 1 - tan x { 2 tan x / (1 - tan^2(x) ]
= [tan x (1 - tan^2(x)) + 2tan x ] / [1 - tan^2(x) - 2tan^2(x) ]
= [ 3tan x - tan^3(x) ] / [1 - 3tan^2(x) ]
= tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ]
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= tan(x + 2x)
= [ tan x + tan(2x) ] / [ 1 - tan(x) tan(2x)]
= [ tan x + { 2 tan x / (1 - tan^2(x) } ] /[ 1 - tan x { 2 tan x / (1 - tan^2(x) ]
= [tan x (1 - tan^2(x)) + 2tan x ] / [1 - tan^2(x) - 2tan^2(x) ]
= [ 3tan x - tan^3(x) ] / [1 - 3tan^2(x) ]
= tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ]
= [ tan x + tan(2x) ] / [ 1 - tan(x) tan(2x)]
= [ tan x + { 2 tan x / (1 - tan^2(x) } ] /[ 1 - tan x { 2 tan x / (1 - tan^2(x) ]
= [tan x (1 - tan^2(x)) + 2tan x ] / [1 - tan^2(x) - 2tan^2(x) ]
= [ 3tan x - tan^3(x) ] / [1 - 3tan^2(x) ]
= tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ]