求证[(cosx)^2]/[1/(tanx/2)-tanx/2]=1/4sin2x
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求证[(cosx)^2]/[1/(tanx/2)-tanx/2]=1/4sin2x
(cosx)^2/[1/tan(x/2) - tan(x/2) ]
=(cosx)^2 tan(x/2) /(1 -[ tan(x/2)]^2)
=(cosx)^2 [ sin(x/2)/cos(x/2)] [cos(x/2)]^2/ { [cos(x/2)]^2 - [sin(x/2)]^2 }
= (cosx)^2 sin(x/2) . cos(x/2) / cosx
= cosx ( 1/2) sinx
= (1/4)sin2x
=(cosx)^2 tan(x/2) /(1 -[ tan(x/2)]^2)
=(cosx)^2 [ sin(x/2)/cos(x/2)] [cos(x/2)]^2/ { [cos(x/2)]^2 - [sin(x/2)]^2 }
= (cosx)^2 sin(x/2) . cos(x/2) / cosx
= cosx ( 1/2) sinx
= (1/4)sin2x
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