求证:2(sin2a+1)/(1+sin2a+cos2a)=tana+1

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求证:2(sin2a+1)/(1+sin2a+cos2a)=tana+1

求证:2(sin2a+1)/(1+sin2a+cos2a)=tana+1
求证:2(sin2a+1)/(1+sin2a+cos2a)=tana+1

求证:2(sin2a+1)/(1+sin2a+cos2a)=tana+1
sin2a=2sinacosa 1=(sina)^2+(cosa)^2
2(sin2a+1)=2(sina+cosa)^2
1+sina2a+cos2a=1+2sinacosa+2(cosa)^2-1=2cosa(sina+cosa)
2(sin2a+1)/(1+sin2a+cos2a)=[2(sina+cosa)^2]/[2cosa(sina+cosa)]
=(sina+cosa)/cosa=tana+1

2(sin2a+1)/(1+sin2a+cos2a)
=2(2sinacosa+sin^2 a+cos^2 a)/(1+2sinacosa+2cos^2 a-1)
=2(sina+cosa)^2/2cosa(sina+cosa)
=(sina+cosa)/cosa
=tana+1