(1/cos^280-3/cos^210)*(1/cos20)=(1/cos^2(90-10)-3/cos^210)(1/cos20)=(1/sin^210-3/cos^210)*(1/cos20)=(cos^210-3sin^210)/(sin^210*cos^210)*(1/cos20)=(4cos^210-3)/(sin^210*cos^210)*(1/cos20)=(2cos20-1)/(...
﹙1/cos²80-3/cos²10﹚·1/cos20
1个回答
相关问题
-
(sin50°(1+√3tan10°)-cos20°)/cos80°√(1-cos20°
-
化简sin50°(1+√3tan10°)-cos20°/cos80°√(1-cos20°)
-
(1/cos80平方-3/cos10平方)乘以1/cos20怎么做
-
cos20°cos10°-cos70°cos80°
-
sin50度(1+根号3tan10度)-cos20度/cos80度根号1-cos20度=
-
(1除以cos80平方-1除以cos10平方)乘以 1除以cos20
-
(1/cos80°的平方-2/cos80°的平方)1/cos20°
-
三角函数问题1.分子是sin50°(1+根号3倍tan10°)-cos20° 分母是cos80°根号下1-cos20°
-
:(sin20)^2+(cos80)^2+根号3cos80cos20=?
-
cos40(1+根号3*cot80)求值怎么从2cos40cos50/sin80 到 (cos10+cos90)/cos