tanθ+secθ =(sinθ/cosθ)+1/cosθ =(1+sinθ)/cosθ =[sin(θ/2)+cos(θ/2)]^2/[(cos(θ/2))^2-(sin(θ/2))^2] =[sin(θ/2)+cos(θ/2)]/[cos(θ/2)-sin(θ/2)] =[1+tan(θ/2)]/[1-1·tan(θ/2)] =[tan(π/4)+tan(θ/2)]/[1-tan(π/4)tan(θ/2)] =tan(π/4+θ/2)。
提示:令θ/2+π/4=t,利用万能公式。
用半角公式， 右边=[1-cos(θ+π/2)]/sin(θ+π/2) =[1+sinθ]/cosθ =secθ+tanθ=左边。
左边=(sinθ/cosθ)+(1/cosθ)
=(sinθ+1)/cosθ
=[sin(θ/2)+cos(θ/2)]^2/[cos^2(θ/2)-sin^2(θ/2)]
=[sin(θ/2)+cos(θ/2)]^2/{[cos(θ/2)+sin(θ/2)]*[cos(θ/2)-sin(θ/2)]}
=[sin(θ/2)+cos(θ/2)]/[cos(θ/2)-sin(θ/2)]
右边=[tan(θ/2)+tan(π/4)]/[1-tan(θ/2)*tan(π/4)]
=[tan(θ/2)+1]/[1-tan(θ/2)]
=[sin(θ/2)+cos(θ/2)]/[cos(θ/2)-(sinθ/2)]
所以：左边=右边.
答：用万能代换公式直接可得,tanx=2时,cos2x+sin2x=[1-(tanx)^2]/[1+(tanx)^2]+2tanx/[1+(tanx)^2]=[(1...详情>>
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