Ответы на вопрос:
1)sin²2a-4sin²a=4sin²acos²a-4sin²a=4sin²a(cos²a-1)=4sin²a*(-sin²a)= =-4sin^4a 2)sin²2a-4+4sin^4a=sin²2a-4(1-sin^4)a=sin²2a-4(1-sin²a)(1+sin²a)= =4sin²acos²a-4*cos²a*(1+sin²a)=4sin²acos²a-4cos²a-4cos²asin²a=-4cos²a 3)-4sin^4a/(-4cos²a)=tg²a*sin²a
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