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

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Answer 1 · 2018-05-23T05:48:09

$L H S = \frac{sec 8 α - 1}{sec 4 α - 1}$ $LHS=(sec8alpha-1)/(sec4alpha-1)$

$= \frac{\frac{1}{cos 8 α} - 1}{\frac{1}{cos 4 α} - 1}$ $=(1/(cos8alpha)-1)/(1/(cos4alpha)-1)$

$= \frac{\frac{1 - cos 8 α}{cos 8 α}}{\frac{1 - cos 4 α}{cos 4 α}}$ $=((1-cos8alpha)/(cos8alpha))/((1-cos4alpha)/(cos4alpha))$

$= \frac{2 {sin}^{2} 4 α cos 4 α}{2 {sin}^{2} 2 α cos 8 α}$ $=(2sin^2 4alphacos4alpha)/(2sin^2 2 alphacos8alpha)$

$= \frac{sin 8 α \cdot sin 4 α}{2 {sin}^{2} 2 α cos 8 α}$ $=(sin 8alpha*sin4alpha)/(2sin^2 2 alphacos8alpha)$

$= \frac{sin 8 α \cdot 2 sin 2 α cos 2 α}{2 {sin}^{2} 2 α cos 8 α}$ $=(sin 8alpha*2sin2alphacos2alpha)/(2sin^2 2 alphacos8alpha)$

$= \frac{tan 8 α}{\frac{sin 2 α}{cos 2 α}}$ $=(tan 8alpha)/((sin 2 alpha)/(cos 2alpha)$

$= \frac{tan 8 α}{tan 2 α} = R H S$ $=(tan 8alpha)/(tan 2 alpha)=RHS$