sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)

化简并计算sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌) 求值tan(kπ+π/6)cos(kπ-4π/3),k属于z 化简sin(-2/3pai+kpai)cos(pai/6+kpai)tan(pai/4+kpai),k属于z sina+cosa=-3/根号5,且|Sina|>|cos a|,求tan(a/2) 化简:sin(3丌-a)/tan(a-5丌)x tan(-a+2丌)/cot(丌-a)x cos(4丌-a)/sin(- 已知sina(2a+b)=3sinb,a不等于kπ+π/2,a+b不等于kπ+π/2,k为正整数,求证：tan(a+b) 怎么证明tan(π/(2k+1))×tan（2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1 有人懂不,已知tan(3.14-A)=-3求：1、4sinA-2cosA/5cosA+3sinA的值2、sinA*cos 已知sin(π-α)-cos(-α)=1/5,求tan[(2k+1)π+α]+cot[(2k+1)π-α](k属於Z)的 已知3*sina=2*sin(2*a+b),a+b,ak*pai+pai/2,k属于整数,求证tan(a+b)=5*ta 已知COS θ=-12/13,θ属於(丌,3丌/2),求SIN( θ-丌/4),COS( θ-丌/4)及TAN( θ-丌 1.已知tan=-4/3,求2(sina)^2+sinacosa-3(cos)^2