急求一道三角函数题答案

分两步
1.tan1*tan2*tan3.....tan88*tan89
=(tan1*tan89)*(tan2*tan88).....(tan44*tan46)*tan45
=1*1*1*1*1....*1*1=1

2.sin^2(1)+sin^2(2)+....sin^2(89)+sin^2(90)
=[sin^2(1)+sin^2(89)]+[sin^2(2)+sin^2(87)]+.......[sin^2(44)+sin^2(46)]+sin^2(45)+sin^2(90)
=1+1+1+....+1+1/2+1=45.5

所以答案是1/45.5=2/91

tan1°*tan2°*tan3°*...*tan89°/(sin1°+sin2°+sin3°+...+sin89°)=2cos1°/sin90°+sin89°-sin1°=2sin89°/(2sin45°cos45°+2sin44°cos45°)=(√2)cos(89°/2)/cos(1/2)