证明(1+secx+tanx)/(1+secx-tanx)=(1+sinx)/cosx左边=(cosx/cosx+1/cosx+sinx/cosx)/(cosx/cosx+1/cosx-sinx/cosx)=(cosx+1+sinx)/(cosx+1-sinx)=(cosx+1+sinx)^2/[(cosx+1+sinx)(cosx+1-sinx)]=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²
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证明(1+secx+tanx)/(1+secx-tanx)=(1+sinx)/cosx左边=(cosx/cosx+1/cosx+sinx/cosx)/(cosx/cosx+1/cosx-sinx/cosx)=(cosx+1+sinx)/(cosx+1-sinx)=(cosx+1+sinx)^2/[(cosx+1+sinx)(cosx+1-sinx)]=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²
证明(1+secx+tanx)/(1+secx-tanx)=(1+sinx)/cosx
左边=(cosx/cosx+1/cosx+sinx/cosx)/(cosx/cosx+1/cosx-sinx/cosx)
=(cosx+1+sinx)/(cosx+1-sinx)
=(cosx+1+sinx)^2/[(cosx+1+sinx)(cosx+1-sinx)]
=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²x+2*cosx+1-sin²x)
=(cosx*sinx+cosx+sinx)/[(cosx+1)*cosx]
=(cosx+1)(sinx+1)/[(cosx+1)*cosx]
=(1+sinx)/cosx
=右边
=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²x+2*cosx+1-sin²x)
=(cosx*sinx+cosx+sinx)/[(cosx+1)*cosx]
这个 上一步怎么转换到下一步的昂
证明(1+secx+tanx)/(1+secx-tanx)=(1+sinx)/cosx左边=(cosx/cosx+1/cosx+sinx/cosx)/(cosx/cosx+1/cosx-sinx/cosx)=(cosx+1+sinx)/(cosx+1-sinx)=(cosx+1+sinx)^2/[(cosx+1+sinx)(cosx+1-sinx)]=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²
所以说你给的推导是错误的,分子少了一个+1,否则你无法通过你给的那个式子来推出接下来的两部.
接着上面的推导就可以得到以下的答案了.
=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²x+2*cosx+1-sin²x)
=(1+2*sinx*cosx+2*cosx+2*sinx+1)/)/(cos²x+2*cosx+cos²x)
=(cosx*sinx+cosx+sinx+1)/[(cosx+1)*cosx]
=(cosx+1)(sinx+1)/[(cosx+1)*cosx]
=(1+sinx)/cosx