Show prove the below identity? #1/cos290 + 1/(sqrt3sin250) = 4/sqrt3 #

Question

#1/cos290 + 1/(sqrt3sin250) -= 4/sqrt3#

4156 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-11T03:51:39

#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{cancel(sin40^@)})/cancel((sin40^@))]=4/sqrt3=RHS#

NOTE that #cos(360-A)^@=cosA and sin(180+A)^@=-sinA#

Answer 2 · 2018-05-11T03:58:49

#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#