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Question
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
Solution
`tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
LHS = `tan x + tan (π/3 + x) - tan(π/3 - x)`
`"LHS" = tan x + ((tan (π/3) + tan x)/(1 - tan x tan (π/3))) - ((tan (π/3) - tan x)/(1 + tan x tan (π/3)))`
We know that,
`tan (A + B) = ((tan A + tan B)/(1 - tan A tan B))` and
`tan (A _ B) = ((tan A - tan B)/(1 + tan A tan B))`
So,
`"LHS" = tan x + ((sqrt3 + tan x)/(1 - sqrt3 tan x)) - ((sqrt3 - tan x)/(1 + sqrt3 tan x))`
`"LHS" = tan x + (((1 + sqrt3 tan x)(sqrt3 + tan x) - (1 - sqrt3 tan x)(sqrt3 - tan x))/((1 - sqrt3tan x)(1 + sqrt3 tanx)))`
Simplify and cancel the similar terms of different sign in the above expression
we get,
`"LHS" = tan x + ((0 + 6tan x + 2tan x + 0)/(1 - 3tan^2x))`
`"LHS" = tan x + ((8tan x)/(1 - 3tan^2x))`
`"LHS" = (tan x (1 - 3tan^2x) + 8tan x)/(1 - 3tan^2x)`
`"LHS" = (tan x - 3tan^3x + 8tan x)/(1 - 3tan^2x)`
`"LHS" = (9tan x - 3tan^3x)/(1 - 3tan^2x)`
`"LHS" = 3((3tan x - tan^3x)/(1 - 3tan^2x))`
`"LHS" = 3 tan 3x ...{tan 3x = (3tanx - tan^3x)/(1 - 3tan^2x)`
RHS = 3 tan 3x
Hence proved.
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