分类：教育/科学-学习帮助
问题：化简题化简[cos40+sin50(1+根号3*tan10)]/ 
化简   [c40+sin50(1+根号3  *tan10)]/[sin70(1+cos40)] 省略了度
回答：[c40+sin50(1+√3 *tan10)]/[sin70(1+cos40)] =[cos40+sin50(1+√3 *sin10/cos10]/[sin70(1+cos40)] ={cos40+sin50[(cos10+√3 *sin10)/cos10]}/[sin70(1+cos40)] ={cos40+2sin50[(1/2cos10+√3/2 *sin10)/cos10]}/[sin70(1+cos40)] ={cos40+2sin50[(sin30cos10+cos30*sin10)/cos10]}/[sin70(1+cos40)] ={cos40+2sin50[sin40)/cos10]}/[sin70(1+cos40)] ={cos40+[2cos40sin40)/cos10]}/[sin70(1+cos40)] ={cos40+[sin80)/cos10]}/[sin70(1+cos40)] ={cos40+[cos10)/cos10]}/[sin70(1+cos40)] ={cos40+1}/[sin70(1+cos40)] =1/sin70