求证以下恒等式1. 1-2csc^2A=ctg^4A-csc^4A2.1/cos80-根号3/sin80＝4（80为角度制）
热心网友
1. 1-2csc^2A=ctg^4A-csc^4A左=1-2(1+cot^2A)=-1-2ctg^2A右=ctg^4A-csc^4A=ctg^4A-(1+ctg^2A)^2=ctg^4A-1-2ctg^2A-ctg^4A=-1-2ctg^2A左=右2.(1/cos80)-(√3/sin80)=(sin80-√3cos80)/sin80cos80=2[(1/2)sin80-(√3/2)cos80]/sin80cos80=4sin(80-60)/2sin80cos80=4sin20/sin160=4sin20/sin20=4
热心网友
1.证:(ctgA)^4-(cscA)^4=(cosA)^4/(sinA)^4-1/(sinA)^4=[(cosA)^4-1]/(sinA)^4=[(cosA)^2+1]*[(cosA)^2-1]/(sinA)^4=-[(cosA)^2+1]*(sinA)^2/(sinA)^4=-[2-(sinA)^2]/(sinA)^2=1-2/(sinA)^2=1-2(cscA)^22.证:1/cos80-√3/sin80[本题略去角度的单位:°]=(sin80-√3cos80)/(sin80cos80)=2(1/2*sin80-√3/2*cos80)/(1/2*2sin80cos80)=4sin(80-60)/sin160=4sin20/sin20=4
热心网友
1.左＝ctg^4A-csc^4A＝cos^4A/sin^4A-1/sin^4A=(cos^4A-1)/sin^4A=(cos^2A+1)(cos^2A-1)/sin^4A==(cos^2A+1)(-sin^2A)/sin^4A=-(cos^2A+1)/sin^2A＝-ctg^2A-csc^2A=-(csc^2A-1)-csc^2A=1-2csc^2A＝右2.1/cos80-根号3/sin80＝（sin80－根号3*cos80）/cos80*sin80=2sin(80-60)/cos80*sin80＝4sin20/sin160=4sin20/sin20=4