作业帮tan(x-π/12)/tan(x+π/12)=1/3,则tanx的值为
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tan(x-π/12)/tan(x+π/12)=1/3,则tanx的值为
tan(x-π/12)/tan(x+π/12)=1/3,则tanx的值为
tan(x-π/12)/tan(x+π/12)=1/3,则tanx的值为
设 a = tan(Pi/12) = tan( Pi/3 - Pi/4) = .= 2- √3
tan(x-π/12)/tan(x+π/12)=1/3 => 3(tanx - a)(1 - a*tanx) - (tanx + a)(1 + a* tanx) = 0
令 y= tanx
整理：y^2 - [ (1 + a^2) / (2a)] * y + 1 = 0
a=2- √3,(1 + a^2) / (2a) = 2 ,=> (y-1)^2 = 1
y=1,即 tanx =1
tan(-x+π/4)
tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x
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