Asked by TeeKay
Find tan2A when tanA = 1/3 and π<A<3π/2.
A) 4/5
B) 5/4
C) 3/4
D) 3/5
A) 4/5
B) 5/4
C) 3/4
D) 3/5
Answers
Answered by
Reiny
So A in in quadrant III
tan 2A = 2tanA/(1 - tan^2 A)
= 2(1/3)/(1 - 1/9)
= (2/3) / (8/9)
= (2/3)(9/8)
= 18/24 = 3/4
tan 2A = 2tanA/(1 - tan^2 A)
= 2(1/3)/(1 - 1/9)
= (2/3) / (8/9)
= (2/3)(9/8)
= 18/24 = 3/4
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