①化简：tan(-35π/6）×sin(-46π/3)-cos(37π/6)×tan(55π/6)
问题描述：
①化简：tan(-35π/6）×sin(-46π/3)-cos(37π/6)×tan(55π/6)
②已知tan(5π＋a)=m(m≠±1),求sin(a-3π)+cos(π－a)/sin(-a)+cos(2π－a)
③已知sin(π/6+a)=1/4,求sin(5π/6－a)+cos^2（7π/6＋a)的值
答
1=tan(π/6)*sin(2π/3)-cos(π/6)*tan(π/6)=(1/√3)*(√3/2)-(√3/2)*(1/√3)=1/2-1/2=02、tan(5π＋a)=m,则tan(a)=m[sin(a-3π)+cos(π－a)]/[sin(-a)+cos(2π－a)]=[-sin(a)-cos(a)]/[-sin(a)+cos(a)]=[sin(a)+c...