tan A +tan(60+A) - tan (60-A) = what 1) 3tan3A 2) tan3A 3)cot 3A 4) sin3A

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

Prove that following identities:

$t a n A + t a n (60^{\circ} + A) - t a n (60^{\circ} - A) = 3 t a n 3 A$

tan (60^{o} - A) tan A tan (60^{o} + A) = tan 3 A

tan A + tan (60^{\circ} + A) + tan (120^{\circ} + A) = 3 tan 3 A

(tan 4 A + tan 2 A) (1 - {tan}^{2} 3 A {tan}^{2} A) = 2 tan 3 A {sec}^{2} A

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