求证(1+sin2α)/(2cos²α+sin2α)=1/2tanα+1/2
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求证(1+sin2α)/(2cos²α+sin2α)=1/2tanα+1/2
求证(1+sin2α)/(2cos²α+sin2α)=1/2tanα+1/2
求证(1+sin2α)/(2cos²α+sin2α)=1/2tanα+1/2
1+sin2a = (sina + cosa)^2
2cos^2a + sin2a = 2cosa(cosa + sina)
原式
=( sina + cosa)/2cosa
= 1/2tana + 1/2