Question

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

Answer 1

`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`