let sin^-1 (3/5) = Ų
then sinŲ = 3/5 and cosŲ = 4/5
so you want tan 2Ų
= 2sinŲcosŲ/(2cos^2Ų - 1)
= 2(3/5)(4/5)/(32/25 - 1)
=(24/25) / (7/25)
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