sec8A-1/sec4A-1=tan8A/tan2A

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*

Thanks very much

To solve this equation involving trigonometric identities, we'll need to simplify both sides of the equation separately.

Let's start with the left side:

sec(8A - 1) / sec(4A - 1)

Using the reciprocal identity, we can rewrite sec as 1/cos:

(1/cos(8A - 1)) / (1/cos(4A - 1))

Next, apply the division rule for fractions by multiplying the first fraction by the reciprocal of the second fraction:

(1/cos(8A - 1)) * (cos(4A - 1)/1)

The cosines in the numerator and denominator can be simplified by using the difference identity for cosine:

(1/cos(8A - 1)) * (cos(4A)cos(1) + sin(4A)sin(1)) / 1

Now, let's move to the right side of the equation:

tan(8A) / tan(2A)

Using the sine and cosine identities for tangent, we can rewrite tan as sin/cos:

(sin(8A)/cos(8A)) / (sin(2A)/cos(2A))

Again, apply the division rule for fractions:

(sin(8A)/cos(8A)) * (cos(2A)/sin(2A))

Now, we can use the double-angle identities for sine and cosine:

(2sin(4A)cos(4A)/cos(8A)) * (cos(2A)/(2sin(2A)cos(2A)))

Cancel out common factors:

(sin(4A)/cos(8A)) * (1/sin(2A))

Now, we'll simplify further:

(sin(4A)/cos(8A)) / (sin(2A)/1)

Apply the division rule again:

(sin(4A)/cos(8A)) * (1/sin(2A))

Now that we have both sides of the equation simplified, they should be equal:

(1/cos(8A - 1)) * (cos(4A)cos(1) + sin(4A)sin(1)) / 1 = (sin(4A)/cos(8A)) * (1/sin(2A))

Both sides are equivalent, so the given equation is true.