求第8题具体过程quq
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sec x+tanx=1/cosx+sinx/cosx=(1+sinx)/cosx
tan(π/4 +x/2)=[tanπ/4+tan(x/2)]/[1-tan(x/2)]
=[1+tan(x/2)]/[1-tan(x/2)]
=[(cos(x/2)+sin(x/2)]/[(cos(x/2)-sin(x/2)]
=[(cos(x/2)+sin(x/2)]²/[(cos(x/2)-sin(x/2)][(cos(x/2)+sin(x/2)]
=[cos²(x/2)+sin²(x/2)+2cos(x/2)sin(x/2)]/[cos²(x/2)-sin²(x/2)]
=(1+sinx)/cosx
所以,sec x+tanx=tan(π/4 +x/2)
tan(π/4 +x/2)=[tanπ/4+tan(x/2)]/[1-tan(x/2)]
=[1+tan(x/2)]/[1-tan(x/2)]
=[(cos(x/2)+sin(x/2)]/[(cos(x/2)-sin(x/2)]
=[(cos(x/2)+sin(x/2)]²/[(cos(x/2)-sin(x/2)][(cos(x/2)+sin(x/2)]
=[cos²(x/2)+sin²(x/2)+2cos(x/2)sin(x/2)]/[cos²(x/2)-sin²(x/2)]
=(1+sinx)/cosx
所以,sec x+tanx=tan(π/4 +x/2)
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