tan(-150°)cos(-570°)cos(-1140°)/cot(-240°)sin（-690°)

tan(-150°)cos(-570°)cos(-1140°)/cot(-240°)sin（-690°) tan(-150°)cos(-210°)cos(-420°)tan(-600°)/sin(-1050°) cos150°×cos(-570°)×tan(-330°)/cos(-420°)×sin(-690°) tan(-2010°)+cos(-79π/6) 计算 tan 30°+cos 30°/tan 45°-cos 60° 等于多少 cos^(50°+α)+cos^(40°-α)-tan(30°-α)*tan(60°+α) 计算 tan 30°+cos 30°/tan 45°-cos 60= 求证cos（720°+α）（2/cosα+tanα）（1/cosα-2tanα）=2cosα-3tanα 计算cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπtan10°t 计算（1）cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6cosπ/3-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6[tan(-150°)cos（-210°）cos420°]/cot(-600°)sin（-1050°）cos25/6π+cos25/3π+tan(-25/4π)5sinπ/2+2cos0-3sin3/2π+10cosπt tan675°+cos765°-tan(-300°)+cos(-690°) cos 60°+sin 45°-tan 30°=? cos tan sin 30° 45° 60°是多少 化简,cos²(-α)-tan(360°+α)/sin(-α) 化简cos^2(-α)-[tan(360°+α)/sin(-α)] 53°的直角三角形各个tan,cos,sin值是多少 高中数学中sin、cos、tan（90°）分别是多少 (tan-150度)cos(-570度)cos(-1140度)tan(-240度)/sin(-690度）等于多少