Question

please solve this question ..........please......​

Answer 1

we know:- tan(A + B) = [{tanA + tanB}/{1 - tanA.tanB}]

therefore,

tanx + {tanx + tanπ/3}/{1 - tanx.tanπ/3} + {tanx + tan2π/3}/{1 - tanx.tan2π/3} = 3

=> tanx + {tanx + √3}/{1 - √3tanx} + {tanx - √3}/{1 + √3tanx} = 3

=>  = 3

=> tanx + {tanx + √3tan²x + √3 + 3tanx + tanx - √3tan²x - √3 + 3tanx}/{1 - 3tan²x} = 3

=> tanx + {8tanx}/{1 - 3tan²x} = 3

=> {tanx - 3tan³x + 8tanx}/{1 - 3tan²x} = 3

=> {9tanx - 3tan³x}/{1 - 3tan²x} = 3

=> 3{tanx - 3tan³x}/{1 - 3tan²x} = 3

we know:- tan3A = {tanA - 3tan³A}/{1 - 3tan²A}

therefore, tan3x = 3/3

=> tan3x = 1 [hence, proved]

Answer 2

Answer:

=> we know:- tan(A + B) = [{tanA + tanB}/{1 - tanA.tanB}]

therefore,

tanx + {tanx + tanπ/3}/{1 - tanx.tanπ/3} + {tanx + tan2π/3}/{1 - tanx.tan2π/3} = 3

=> tanx + {tanx + √3}/{1 - √3tanx} + {tanx - √3}/{1 + √3tanx} = 3

=>  = 3

=> tanx + {tanx + √3tan²x + √3 + 3tanx + tanx - √3tan²x - √3 + 3tanx}/{1 - 3tan²x} = 3

=> tanx + {8tanx}/{1 - 3tan²x} = 3

=> {tanx - 3tan³x + 8tanx}/{1 - 3tan²x} = 3

=> {9tanx - 3tan³x}/{1 - 3tan²x} = 3

=> 3{tanx - 3tan³x}/{1 - 3tan²x} = 3

we know:- tan3A = {tanA - 3tan³A}/{1 - 3tan²A}

therefore, tan3x = 3/3

=> tan3x = 1 [hence, proved]

Step-by-step explanation:

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