Question

sec8A-1/sec4A-1=tan8A/tan2A

Answer 1

to prove

(sec8A-1)/(sec4A-1) = tan8A/tan2A

Not sure how to do this with sec and tan only, but using sin and cos, we have

sec8A-1 = (1-cos8A)/cos8A
= (1-(2cos^2(4A)-1))/cos8A
= 2(1-cos^2(4A))/cos8A

sec4A-1 = (1-cos4A)/cos4A

so, the LS is
2cos4A(1-cos^2(4A))/(2cos^2(4A)-1)*(1-cos4A)
= 2cos4A(1+cos4A)/cos8A

on the RS, we use half-angle formula

tan2A = sin4A/(1+cos4A)
and
tan8A = sin8A/cos8A = 2sin4Acos4A/cos8A

divide to get

2sinAcos4A/cos8A * (1+cos4A)/sin4A
= 2cos4A(1+cos4A)/cos8A

and LS = RS
*whew*

Answer 2

Thanks very much