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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
人气:182 ℃ 时间:2020-05-20 00:46:01
解答
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)=√...
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