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Prove that ππtanx+tan(π3+x)-tan(π3-x)=3tan3x - Mathematics

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प्रश्न

Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`

योग
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उत्तर

`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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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.2 [पृष्ठ ३६]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.2 | Q 4 | पृष्ठ ३६

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