1+tanα/1-tanα=2006则1/cosα+tan2α=?

wangpeng19862022-10-04 11:39:541条回答

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妍欢儿 共回答了16个问题 | 采纳率87.5%
(1+tanα)/(1-tanα)=2006
1+tanα=2006(1-tanα)
tanα=2005/2007
cos2α=(1-tan^2α)/(1+tan^2α)
=(1-2005^2/2007^2)/(1+2005^2/2007^2)
tan2α=2tanα/(1-tan^2α)
=(2*2005/2007)/(1-2005^2/2007^2)
1/cos2α+tan2α
=1/[(1-2005^2/2007^2)/(1+2005^2/2007^2)]+(2*2005/2007)/(1-2005^2/2007^2)
=(1+2005^2/2007^2)/(1-2005^2/2007^2)+(2*2005/2007)/(1-2005^2/2007^2)
=(1+2005^2/2007^2+2*2005/2007)/(1-2005^2/2007^2)
=(1+2005/2007)^2/[(1-2005/2007)(1+2005/2007)]
=(1+2005/2007)/(1-2005/2007)
=(4012/2007)/(2/2007)
=4012/2
=2006
1/cos2α+tan2α
=1/[(1-tan^2α)/(1+tan^2α)]+2tanα/(1-tan^2α)
=(1+tan^2α)/(1-tan^2α)+2tanα/(1-tan^2α)
=(1+tan^2α+2tanα)/(1-tan^2α)
=(1+tanα)^2/(1-tan^2α)
=(1+tanα)^2/[(1-tanα)(1+tanα)]
=(1+tanα)/(1-tanα)
=(1+tanα)/(1-tanα)
=2006
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一道有点难度的题,尽快解决已知X=2005A+2004 Y=2005A+2005 Z=2005A+2006则X^2+Y^
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已知X=2005A+2004
Y=2005A+2005
Z=2005A+2006
则X^2+Y^2+Z^2-XY-YZ-XZ的值是?
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apperever 共回答了18个问题 | 采纳率88.9%
X=2005A+2004
Y=2005A+2005
Z=2005A+2006
x-y=-1
y-z=-1
x-z=-2
X^2+Y^2+Z^2-XY-YZ-XZ
=(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2)/2
=[(x-y)^2+(x-z)^2+(y-z)^2]/2
=(1+1+4)/2
=3