证明下列恒等式1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α证明下列恒等式:1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^24、(tanA-tanB)/(cotB-cotA
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![证明下列恒等式1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α证明下列恒等式:1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^24、(tanA-tanB)/(cotB-cotA](/uploads/image/z/1063648-64-8.jpg?t=%E8%AF%81%E6%98%8E%E4%B8%8B%E5%88%97%E6%81%92%E7%AD%89%E5%BC%8F1%E3%80%81cos%5E2%CE%B1%2B2sin%5E2%CE%B1%2Bsin%5E2%CE%B1tan%5E2%CE%B1%3D1%2Fcos%5E2%CE%B1%E8%AF%81%E6%98%8E%E4%B8%8B%E5%88%97%E6%81%92%E7%AD%89%E5%BC%8F%3A1%E3%80%81cos%5E2%CE%B1%2B2sin%5E2%CE%B1%2Bsin%5E2%CE%B1tan%5E2%CE%B1%3D1%2Fcos%5E2%CE%B12%E3%80%81cos%5E2%CE%B1%282%2Btan%CE%B1%29%281%2B2tan%CE%B1%29%3D2%2B5sin%CE%B1cos%CE%B13%E3%80%81%281%2Btan%5E2A%29%2F%281%2Bcot%5E2A%29%3D%5B%281-tanA%29%2F%281-cotA%29%5D%5E24%E3%80%81%28tanA-tanB%29%2F%28cotB-cotA)
证明下列恒等式1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α证明下列恒等式:1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^24、(tanA-tanB)/(cotB-cotA
证明下列恒等式1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α
证明下列恒等式:
1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α
2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα
3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^2
4、(tanA-tanB)/(cotB-cotA)=tanB/cotA
证明下列恒等式1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α证明下列恒等式:1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^24、(tanA-tanB)/(cotB-cotA
1.1+tan^α=(cos^α+sin^α)/cos^α=1/cos^α,
∴左边=cos^α+sin^α+sin^α(1+tan^α)
=1+sin^α/cos^α
=1+tan^α
=1/cos^α=右边.
2.左边=(2cosα+sinα)(cosα+2sinα)
=2cos^α+5sinαcosα+2sin^α
=2+5sinαcosα=右边.
3.左边=(1/cos^A)/(1/sin^A)=(sinA/cosA)^=tan^A,
右边={[(cosA-sinA)/cosA]/[(sinA-cosA)/sinA]}^
=(-sinA/cosA)^
=tan^A=左边.
4.左边=(tanA-tanB)/[1/tanA-1/tanB]
=-tanAtanB,
右边=tanAtanB≠左边.
请检查第4题题目.