(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)=sinα乘cosα麻烦帮忙证明下,
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(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)=sinα乘cosα麻烦帮忙证明下,
(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)=sinα乘cosα
麻烦帮忙证明下,
(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)=sinα乘cosα麻烦帮忙证明下,
(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)
=cotα(tanα+tan^2α+1)/(1+tan^2α+tanα)-cotα/(1+tan^2α)
=cotα(1-1/(1+tan^2α))
=cotα*tan^2α/(1+tan^2α)
=tanα/(1+tan^2a)
=sinα/cosα*(1+sin^2α/cos^2α)
=sinαcosα/(cos^2α+sin^2α)
=sinα*cosα=右边
原式左边=(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)
=(1+tanα+1/tanα)/(1+tan^2α+tanα)-1/[tanα(1+tan^2α)]
=(tanα+tan^2α+1)/[tanα(1+tan^2α+tanα)]-1/[tanα(1+tan^2α)]
=1/[tanα-1/[tanα(1+tan^2α)]...
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原式左边=(1+tanα+cotα)/(1+tan^2α+tanα)-cotα/(1+tan^2α)
=(1+tanα+1/tanα)/(1+tan^2α+tanα)-1/[tanα(1+tan^2α)]
=(tanα+tan^2α+1)/[tanα(1+tan^2α+tanα)]-1/[tanα(1+tan^2α)]
=1/[tanα-1/[tanα(1+tan^2α)]
=(1+tan^2α-1)/[tanα(1+tan^2α)]
=tanα/(1+tan^2α)…………因为sin2α=2tanα/(1+tan^2α)
=(sin2α)/2
=sinαcosα
=右边
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