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

Question

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

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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

$1/cos290 + 1/(sqrt3sin250)$

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

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

$=[(sqrt3cos20-sin20)/(sqrt3sin20cos20)]$

$=2/sqrt3[(sqrt3/2cos20-1/2sin20)/(sin20cos20)]$

$=4/sqrt3[(sin60cos20-cos60sin20)/(2sin20cos20)]$

$=4/sqrt3[sin(60-20)/(2sin20cos20)]$

$=4/sqrt3[sin40/sin40]$

$= 4/sqrt3$