显示证明以下身份? 1 / cos290 + 1 /(sqrt3sin250)= 4 / sqrt3

显示证明以下身份? 1 / cos290 + 1 /(sqrt3sin250)= 4 / sqrt3
Anonim

#LHS = 1 /(cos290 ^ @)+ 1 /(sqrt3sin250 ^ @)#

#= 1 /(COS(360-70)^ @)+ 1 /(sqrt3sin(180 + 70)^ @)#

#= 1 /(cos70 ^ @) - 1 /(sqrt3sin70 ^ @)#

#=(sqrt3sin70 ^ @ - cos70 ^ @)/(sqrt3sin70 ^ @ cos70 ^ @)#

#= 1 / sqrt3 (2- {sqrt3sin70 ^ @ - cos70 ^ @})/(2sin70 ^ @ cos70 ^ @)#

#= 1 / sqrt3 (2 * 2 {sin70 ^ @ *(sqrt3 / 2)^ -cos70 @ *(1/2)})/(sin140 ^ @)#

#= 1 / sqrt3 (4- {sin70 ^ @ * cos30 ^ @ - cos70 ^ @ * sin30 ^ @})/(SIN(180-40)^ @)#

#= 1 / sqrt3 (4- {SIN(70-30)^ @})/(sin40 ^ @) = 1 / sqrt3 (4- {取消(sin40 ^ @)})/取消((sin40 ^ @) ) = 4 / sqrt3 = RHS#

注意 #cos(360-A)^ @ = cosA和sin(180 + A)^ @ = - sinA#

#1 / cos290 + 1 /(sqrt3sin250)#

#= 1 / cos(270 + 20)+ 1 /(sqrt3sin(270-20))#

#= 1 / sin20 - 1 /(sqrt3cos20)#

#= (sqrt3cos20-sin20)/(sqrt3sin20cos20)#

#= 2 / sqrt3 (sqrt3 / 2cos20-1 / 2sin20)/(sin20cos20)#

#= 4 / sqrt3 (sin60cos20-cos60sin20)/(2sin20cos20)#

#= 4 / sqrt3 sin(60-20)/(2sin20cos20)#

#= 4 / sqrt3 sin40 / sin40#

#= 4 / sqrt3#