Question

Proove sec 8x-1/sec 4x-1=tan8x/tan2x

Answer 1

LHS=(sec8x-1)/(sec4x-1)

we know,
cos2x=cos^2x-sin^2x
=2cos^2x-1

it means ,
cos8x=2cos^2(4x)-1
1-cos8x=2-2cos^2(4x)=2(1-cos^2(4x))

now,
(sec8x-1)/(sec4x-1)=(1-cos8x).cos4x/(1-cos4x).cos8x
=2(1-cos4x)(1+cos4x).cos4x/(1-cos4x).cos8x)

=2(1+cos4x).cos4x/cos8x

=2.2cos^2(2x).cos4x/cos8x

{(1+cosx)=2cos^2(x/2)

=4(cos2x)^2 .cos4x.sin8x/sin8x.cos8x

=4tan8x.cos4x.(cos2x)^2/sin8x

=4tan8x.cos4x.(cos2x)^2/4sin2x.cos2x.cos4x. {use formula sin2x=2sinx.cosx

=tan8x/tan2x=RHS