(1+cos4a+sin4a)/(1-cos4a+sin4a)化简为?
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![(1+cos4a+sin4a)/(1-cos4a+sin4a)化简为?](/uploads/image/z/6864923-11-3.jpg?t=%281%2Bcos4a%2Bsin4a%29%2F%281-cos4a%2Bsin4a%29%E5%8C%96%E7%AE%80%E4%B8%BA%3F)
(1+cos4a+sin4a)/(1-cos4a+sin4a)化简为?
(1+cos4a+sin4a)/(1-cos4a+sin4a)化简为?
(1+cos4a+sin4a)/(1-cos4a+sin4a)化简为?
解析:(1+cos4a+sin4a)/(1-cos4a+sin4a)
=(1+2cos^22a-1+2sin2acos2a)/(1-1+2sin^22a+2sin2acos2a)
=(2cos^22a+2sin2acos2a)/(2sin^22a+2sin2acos2a)
=cos2a(cos2a+sin2a)/sin2a(cos2a+sin2a)
=cos2a/sin2a
=1/tan2a
打字不易,