Question

TanAtan(A+60)+tanAtan(A-60)+tan(A+60)tan(A-60)=3

Answer 1

There is a small mistake in your question. The RHS is -3

tanA tan(A + 60°) + tanA tan(A - 60°) + tan(A - 60°) tan(A + 60°) = - 3

let's apply (to both tan(A + 60°) ) the addition formula tan(α + β) =
(tan α + tan β) /(1 - tan α tan β):

tanA {[tanA + tan(60°)] /[1 - tanA tan(60°)]} + tanA tan(A - 60°) + tan(A -
60°) {[tanA + tan(60°)] /[1 - tanA tan(60°)]} =

let's apply (to both tan(A - 60°) ) the subtraction formula tan(α - β) =
(tan α - tan β) /(1 + tan α tan β):

tanA {[tanA + tan(60°)] /[1 - tanA tan(60°)]} + tanA {[tanA - tan(60°)] /[1 +
tanA tan(60°)]} + {[tanA - tan(60°)] /[1 + tanA tan(60°)]} {[tanA + tan(60°)] /[1 -
tanA tan(60°)]} =

(being tan(60°) = √3)

tanA {(tanA + √3) /[1 - tanA (√3)]} + tanA {(tanA - √3) /[1 + tanA (√3)]} +
{(tanA - √3) /[1 + tanA (√3)]} {(tanA + √3) /[1 - tanA (√3)]} =

{[tanA (tanA + √3)] /[1 - (√3)tanA]} + {[tanA (tanA - √3)] /[1 + (√3)tanA]} +
{[(tanA - √3)(tanA + √3)] /{[1 + (√3)tanA][1 - (√3)tanA]} } =

(letting [1 + (√3)tanA][1 - (√3)tanA] be the common denominator)

{tanA (tanA + √3) [1 + (√3)tanA] + tanA (tanA - √3) [1 - (√3)tanA] + (tanA - √3)(tanA + √3)} /{[1 + (√3)tanA][1 - (√3)tanA]} =

(expanding)

{tanA [tanA + (√3)tan²A + √3 + (√3)²tanA] + tanA [tanA - (√3)tan²A - √3 +
(√3)²tanA] + [tan²A - (√3)²]} /{[1 + (√3)tanA][1 - (√3)tanA]} =

[tan²A + (√3)tan³A + (√3)tanA + 3tan²A + tan²A - (√3)tan³A - (√3)tanA +
3tan²A + tan²A - 3] /{[1 + (√3)tanA][1 - (√3)tanA]} =

(adding like terms together)

(9tan²A - 3) /{[1 + (√3)tanA][1 - (√3)tanA]} =

(expanding the denominator)

(9tan²A - 3) /{1² - [(√3)tanA]²} =

(9tan²A - 3) /[1 - (√3)²tan²A] =

(factoring - 3 out of the numerator)

[- 3 (- 3tan²A + 1)] /(1 - 3tan²A) =

[- 3 (1 - 3tan²A)] /(1 - 3tan²A) =

simplifying to:


- 3


I hope it's helpful