根号3tan(pai/6-θ)tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=?
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/04/28 15:35:01
根号3tan(pai/6-θ)tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=?
根3*tan(pai/6-θ)*tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=?
用三角的恒等变换
根3*tan(pai/6-θ)*tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=?
用三角的恒等变换
tan(π/6-θ) =[ tanπ/6-tanθ]/(1+tanπ/6tanθ)
= [1/√3 - tanθ]/(1+tanθ/√3)
= (1-√3tanθ)/(√3+tanθ)
tan(π/6+θ) =(1+√3tanθ)/(√3-tanθ)
√3tan(π/6-θ)tan(π/6+θ)+tan(π/6-θ)+tan(π/6+θ)
=√3[(1-√3tanθ)/(√3+tanθ)][(1+√3tanθ)/(√3-tanθ)]+(1+√3tanθ)/(√3-tanθ)+ (1-√3tanθ)/(√3+tanθ)
=√3(1-3(tanθ)^2)/(3-(tanθ)^2) +[(1+√3tanθ)(√3+tanθ)+(1-√3tanθ)(√3-tanθ)]/(3-(tanθ)^2)
=[√3(1-3(tanθ)^2) + 2√3+2√3(tanθ)^2 ] /(3-(tanθ)^2)
= √3(3-(tanθ)^2) /(3-(tanθ)^2)
= √3
= [1/√3 - tanθ]/(1+tanθ/√3)
= (1-√3tanθ)/(√3+tanθ)
tan(π/6+θ) =(1+√3tanθ)/(√3-tanθ)
√3tan(π/6-θ)tan(π/6+θ)+tan(π/6-θ)+tan(π/6+θ)
=√3[(1-√3tanθ)/(√3+tanθ)][(1+√3tanθ)/(√3-tanθ)]+(1+√3tanθ)/(√3-tanθ)+ (1-√3tanθ)/(√3+tanθ)
=√3(1-3(tanθ)^2)/(3-(tanθ)^2) +[(1+√3tanθ)(√3+tanθ)+(1-√3tanθ)(√3-tanθ)]/(3-(tanθ)^2)
=[√3(1-3(tanθ)^2) + 2√3+2√3(tanθ)^2 ] /(3-(tanθ)^2)
= √3(3-(tanθ)^2) /(3-(tanθ)^2)
= √3
根号3tan(pai/6-θ)tan(pai/6+θ)+tan(pai/6-θ)+tan(pai/6+θ)=?
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