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Given, 2 tan a = 3 tan b
tan a = 3/2 tan b ---------- (1)
We know that tan(a-b) = (tana - tanb)/(1 + tanatanb)
= (3/2 tanb - tanb)/(1 + 3/2 tanbtanb)
= tanb(3/2 - 1)/(1 + 3/2 tan^2b)
= tanb(1/2)/(sec2b - tan2b + 3/2 tan2^b)
= sinb cosb/(2+sin2^b)
= 2sinb cosb/(4+2sin^2b)
= sin2b/(4+1-cos^2b)
.
= sin2b/(5 - cos2b)
tan a = 3/2 tan b ---------- (1)
We know that tan(a-b) = (tana - tanb)/(1 + tanatanb)
= (3/2 tanb - tanb)/(1 + 3/2 tanbtanb)
= tanb(3/2 - 1)/(1 + 3/2 tan^2b)
= tanb(1/2)/(sec2b - tan2b + 3/2 tan2^b)
= sinb cosb/(2+sin2^b)
= 2sinb cosb/(4+2sin^2b)
= sin2b/(4+1-cos^2b)
.
= sin2b/(5 - cos2b)
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