Q.378. Trigonometric derivation.

Question 378.
Express tan5A in terms of tanA.

Answer 378.
tan5A
= tan(2A + 3A)
= (tan2A + tan3A) / (1 - tan2A tan3A) ... ( 1 )

tan2A + tan3A
= 2tanA / (1 - tan^2 A) + (3tanA - tan^3 A) / (1 - 3tan^2 A)
= (2tanA - 6tan^3 A + 3tanA - 3tan^3 A - tan^3 A + tan^5 A)
    diided by [(1 - tan^2 A)(1 - 3tan^2A]
= (tan^5 A - 10tan^3 A + 5tanA) / [(1 - tan^2 A)(1 - 3tan^2A] ... ( 2 )

1 - tan2A tan3A
= 1 - [2tanA (3tanA - tan^3 A)] / [(1 - tan^2 A)(1 - 3tan^2A]
= (1 - 4tan^2 A + 3tan^4 A - 6tan^2 A + 2tan^4 A) / [(1 - tan^2 A)(1 - 3tan^2A]
= (1 - 10tan^2 A + 5tan^4 A) / [(1 - tan^2 A)(1 - 3tan^2A] ... ( 3 )

Putting results ( 2 ) and ( 3 ) in ( 1 ),
tan5A
= (tan^5 A - 10tan^3 A + 5tanA) / (1 - 10tan^2 A + 5tan^4 A).

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