Prove that: #(sec8theta-1)/(sec4theta-1)=(tan8theta)/(tan2theta)#?

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Answer 1 · 2018-02-20T11:16:02

#LHS=(sec 8theta -1 )/ (sec4theta -1)#

#=(1/(cos8theta) -1 )/ (1/(cos4theta) -1)#

#=((1-cos8theta)cos4theta )/ ((1-cos4theta) cos8theta)#

#=(2sin^2 4thetacos4theta )/ ((2sin^2 2theta) cos8theta)#

#=(2sin4thetacos4thetasin4theta )/ ((2sin^2 2theta) cos8theta)#

#=(sin8theta*2sin2thetacos2theta )/ ((2sin^2 2theta) cos8theta)#

#=(tan8theta*cos2theta )/ (sin2theta)#

#= (tan8theta) /(tan 2theta)=RHS#