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Question
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
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Solution
Given that: tanA = `(1 - cos "B")/sin"B"`
tan2A = `(2tan"A")/(1 - tan^2"A")`
= `(2((1 - cos "B")/sin"B"))/(1 - ((1 - cos "B")/sin"B")^2`
= `(2((2sin^2 "B"/2)/(2sin "B"/2 cos "B"/2)))/(1 - ((2sin^2 "B"/2)/(2sin "B"/2 cos "B"/2))^2` ........`[(because 1 - cos "B" = 2sin^2 "B"/2),(sin"B" = 2sin"B"/2 cos"B"/2)]`
= `(2((sin "B"/2)/(cos "B"/2)))/(1 - ((sin "B"/2)/(cos "B"/2))^2`
= `(2tan "B"/2)/(1 - tan^2 "B"/2)`
= tanB
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