三角函数计算(sinx)^2/(sinx-cosx)-(sinx+cosx)/((tanx)^2-1) 化简求值
这个题目难度如何
人气:439 ℃ 时间:2019-08-17 20:20:28
解答
((tanx)^2-1)
= (sinx)^2/(cosx)^2 - 1
=( (sinx)^2 - (cosx)^2 ) / (cosx)^2
=(sinx-cosx)(sinx+cosx) / (cosx)^2 代入本式子
得(sinx)^2/(sinx-cosx)-(sinx+cosx)/((tanx)^2-1)
= (sinx)^2/(sinx-cosx)- (sinx+cosx) /((sinx-cosx)(sinx+cosx) / (cosx)^2 )
=(sinx)^2/(sinx-cosx)- (cosx)^2 /(sinx-cosx)
= ((sinx)^2 - (cosx)^2 ) /(sinx-cosx)
=((sinx-cosx)(sinx+cosx) /(sinx-cosx)
= sinx+cosx
推荐
猜你喜欢
- do you like apples,bananas,tomatoes or oranges?do you like apples,or bananas,or tomatoes,or...
- 一、用介词填空 7、You can buy this CD ___ a very good price —only 10 yuan.
- The doctor insisted that the patient should be operated on at once帮忙翻译下,
- “地上河”属于黄河的那一段?位于哪两个省级行政区划单位境内?
- 六年级学的分数除法应用题
- _______(adj.)---that you often see or that often happens
- 燕然未勒归无计中的勒是什么含义
- 已知集合A={1,2}B={-1,0,1}C={0,1}求(AUB)UC?怎么算