Find tan2A when tanA = 1/3 and π<A<3π/2.
A) 4/5
B) 5/4
C) 3/4
D) 3/5
tan(2A) = sin(2A) / cos(2A) = 2sinAcosA / (cos^2A - sin^A)
sinA = 1/√10 cos(A) = 3/√10 so we have
tan(2A) = 2(1/√10)(3/√10) / (9/10 - 1/10) = (6/10) /(8/10) = 6/8 = 3/4
