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#