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[tan(5π/4) + tan(5π/12) ] /[ 1- tan(5π/12) ]
=[ -1 + tan(5π/12) ] /[ 1- tan(5π/12) ]
=-[ 1 - tan(5π/12) ] /[ 1- tan(5π/12) ]
=-1
------
tan(π/8)/ { 1 - [tan(π/8)]^2 }
=(1/2) [ 2tan(π/8)/ { 1 - [tan(π/8)]^2 } ]
=(1/2) tan(π/4)
=1/2
[tan(5π/4) + tan(5π/12) ] /[ 1- tan(5π/12) ]
=[ -1 + tan(5π/12) ] /[ 1- tan(5π/12) ]
=-[ 1 - tan(5π/12) ] /[ 1- tan(5π/12) ]
=-1
------
tan(π/8)/ { 1 - [tan(π/8)]^2 }
=(1/2) [ 2tan(π/8)/ { 1 - [tan(π/8)]^2 } ]
=(1/2) tan(π/4)
=1/2
追问
第二题没看懂
追答
tan(π/8+π/8) = 2tan(π/8)/ { 1 - [tan(π/8)]^2 }
tan(π/4)= 2tan(π/8)/ { 1 - [tan(π/8)]^2 }
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