求证:tanα+1÷tan﹙π÷4+α÷2﹚=1÷cosα
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求证:tanα+1÷tan﹙π÷4+α÷2﹚=1÷cosα
求证:tanα+1÷tan﹙π÷4+α÷2﹚=1÷cosα
求证:tanα+1÷tan﹙π÷4+α÷2﹚=1÷cosα
tanα+1÷tan﹙π÷4+α÷2﹚
=tanα+1/tan(π/4+α/2)
=sinα/cosα+cos(π/4+α/2)/sin(π/4+α/2)
=sinα/cosα+[(cos(α/2)-sin(α/2)]/[cos(α/2)+sin(α/2)]
=sinα/cosα+[cos(α/2)-sin(α/2)]^2/{[cos(α/2)]^2-[sin(α/2)]^2}
=sinα/cosα-[1-2sin(α/2)cos(α/2)]/cosα
=sinα/cosα-(1-sinα)/cosα
=1/cosα
1÷tan﹙π÷4+α÷2﹚
=(1-tan(α/2))/(1+tan(α/2))
=(cos(α/2)-sin(α/2))/(cos(α/2)+sin(α/2))
分子分母同时乘cos(α/2) 再利有半角公式化简
=(1+cos(α)-sin(α))/(1+cos(α)+sin(α))
所以tanα+1÷tan﹙π÷4+α÷2﹚
=sinα/co...
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1÷tan﹙π÷4+α÷2﹚
=(1-tan(α/2))/(1+tan(α/2))
=(cos(α/2)-sin(α/2))/(cos(α/2)+sin(α/2))
分子分母同时乘cos(α/2) 再利有半角公式化简
=(1+cos(α)-sin(α))/(1+cos(α)+sin(α))
所以tanα+1÷tan﹙π÷4+α÷2﹚
=sinα/cosα+(1+cos(α)-sin(α))/(1+cos(α)+sin(α))
通分=1÷cosα
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