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

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

2 Answers
May 11, 2018

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

May 11, 2018

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