Simplify And Prove That ?? (Sec8α-1) / (Sec4α-1) = (Tan8α)/(Tan2α)

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Answer 1 · 2018-05-23T05:48:09

$LHS=(sec8alpha-1)/(sec4alpha-1)$

$=(1/(cos8alpha)-1)/(1/(cos4alpha)-1)$

$=((1-cos8alpha)/(cos8alpha))/((1-cos4alpha)/(cos4alpha))$

$=(2sin^2 4alphacos4alpha)/(2sin^2 2 alphacos8alpha)$

$=(sin 8alpha*sin4alpha)/(2sin^2 2 alphacos8alpha)$

$=(sin 8alpha*2sin2alphacos2alpha)/(2sin^2 2 alphacos8alpha)$

$=(tan 8alpha)/((sin 2 alpha)/(cos 2alpha)$

$=(tan 8alpha)/(tan 2 alpha)=RHS$