Доказать:
sin^6a+cos^6a=1-3/4sin^2a
(1+sin2a+cos2a)/(1+sin2a-cos2a)=ctga
(2sina-sin2a)/(2sina+sin2a)=tg^2a/2
sin^6a+cos^6a=1-3/4sin^2a
(1+sin2a+cos2a)/(1+sin2a-cos2a)=ctga
(2sina-sin2a)/(2sina+sin2a)=tg^2a/2
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еще третий номер.