tanA tan(A-60°) + tanA tan(A+60°) + tan(A-60°) tan(A+60°) = -3?
Prove that tanA tan(A-60°) + tanA tan(A+60°) + tan(A-60°) tan(A+60°) = -3
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Prove that tanA tan(A-60°) + tanA tan(A+60°) + tan(A-60°) tan(A+60°) = -3
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tan(A)tan(A - 60°) + tan(A)tan(A + 60°) + tan(A - 60°)tan(A + 60°)
= tan(A)[tan(A) - tan(60°)]/[1 + tan(A)tan(60°)]
+ tan(A)[tan(A) + tan(60°)]/[1 - tan(A)tan(60°)]
+ {[tan(A) - tan(60°)]/[1 + tan(A)tan(60°)]}{[tan(A) + tan(60°)]/[1 - tan(A)tan(60°)]}
= tan(A)[tan(A) - tan(60°)][1 - tan(A)tan(60°)]/[1 - tan²(A)tan²(60°)]
+ tan(A)[tan(A) + tan(60°)][1 + tan(A)tan(60°)]/[1 - tan²(A)tan²(60°)]
+ [tan(A) - tan(60°)][tan(A) + tan(60°)]/[1 - tan²(A)tan²(60°)]
= tan(A)[tan(A) - √(3)][1 - √(3)tan(A)]/[1 - 3tan²(A)]
+ tan(A)[tan(A) + √(3)][1 + √(3)tan(A)]/[1 - 3tan²(A)]
+ [tan(A) - √(3)][tan(A) + √(3)]/[1 - 3tan²(A)]
= tan(A)[4tan(A) - √(3) - √(3)tan²(A)]/[1 - 3tan²(A)]
+ tan(A)[4tan(A) + √(3) + √(3)tan²(A)]/[1 - 3tan²(A)]
+ [tan²(A) - 3]/[1 - 3tan²(A)]
= [9tan²(A) - 3]/[1 - 3tan²(A)]
= -3[1 - 3tan²(A)]/[1 - 3tan²(A)]
= -3
It is not an identity for all real A. The exclusions are a bit complicated. Here is what I get:
A ≠ 30°, 90°, or 150° (mod 180°)