Show that TanA.Tan(60-A).Tan(60+A) = Tan3A
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Answer:
Step-by-step explanation:
TanA. [Tan60 - TanA/1+Tan60.TanA]. [Tan60 + TanA/1 - Tan60.TanA]
= TanA [Tan60 - TanA][Tan60 + TanA] / [1+Tan60.TanA] [1 - Tan60.TanA]
// [Tan60 - TanA][Tan60 + TanA] is in the form (a+b) (a-b). Similarly [1+Tan60.TanA] [1 - Tan60.TanA] is in the form (a+b) (a-b). But (a+b)(a-b) = a² - b²
= TanA [Tan²60 - Tan²A / 1 - Tan²60Tan²A]
//Tan60 = √3.
= TanA ( 3 - Tan²A] / 1 - 3Tan²A
= 3TanA - Tan³A / 1 - 3Tan²A
//Remember Tan3A = 3TanA - Tan³A / 1 - 3Tan²A
= Tan3A
= R.H.S
Hence Proved
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