Question

hey

Prove......!!!

DoNoT SpAm ​

Answer 1

tan a+ tan( 60+a) + tan( 120 + a)

tan a +( tan60 + tana)/( 1- tan60tana)

+ (tan 120 +tana)/( 1- tan120tana)

= tan a+ ( √3+tan a)/( 1-√3tana)

- √3 + tana)/ ( 1+ √3 tana)

= tana( 1- 3 tan^2a) + ( √3+ tana)( 1+√3tana) +(- √3+ tana)( 1-√3tana)/( 1- 3 tan^2 a)

= tana - 3 tan^3a + √3 +3tana +tana + √3 tan^2a - √3 +3 tan a +tana - √3tan^2a)/ 1-3tan^2a

= -3tan^3a+ 9 tana ) 1- 3 tan^2a

= 3(3 tana - tan^3a)/1- 3tan^2a)

= 3 tana

Answer 2

TAKING,

theta=x

the formula for tan(A+B) = (tanA + tanB) / (1 - tanAtanB)

tan (60°+x) = (tan60° + tanx) / (1 - tan60°tanx) = (√3 + tanx) / (1-√3 tanx) [#1]

tan (120° + x) = (tan120° + tanx) / (1 - tan120°tanx) = (-√3 + tanx) / (1+√3 tanx) [#2]

Multiply [#1] by (1+√3 tanx)/(1+√3 tanx) to get (√3tan²x + 4tanx + √3) / (1-3tan²x)

Multiply [#2] by (1-√3 tanx)/(1-√3 tanx) to get (-√3tan²x + 4tanx - √3) / (1-3tan²x)

The sum of those two terms is (8tanx) / (1-3tan²x)

Thus,

tanx+tan(60°+x)+tan(120°+x)

= tanx + (8tanx) / (1-3tan²x)

= tanx(1-3tan²x)/(1-3tan²x) + (8tanx) / (1-3tan²x)

= (9tanx - 3tan³x) / (1-3tan²x)

= 3(3tanx - tan³x) / (1-3tan²x) ← L.H.S.

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On the R.H.S., I will use both the sum formula for tangent shown above, as well as the double angle formula for tangent: tan(2A) = (2tanA)/(1-tan²A)

3tan(3x)

= 3tan(2x+x)

= 3[(tan2x+tanx) / (1-tan2x·tanx)]

= 3 [(2tanx/(1-tan²x) + tanx) / (1 - 2tan²x/(1-tan²x)]

Multiply by (1-tan²x)/(1-tan²x)

3[3tanx - tan³x) / (1 - 3tan²x)]RHS

LHS=RHS

Hence proved