(1+sec 2A) (1+sec4A) (1+sec8A) = tan8AcotA

Answer 1

recall your half-angle formula:

tan(x/2) = sinx/(1+cosx)

Now, we have

1+sec2A = (1+cos2A)/cos2A
= 1/[cos2A/(1+cos2A)]
= 1/[sin2A/(1+cos2A) * cos2A/sin2A]
= 1/(tanA cot2A)
= tan2A/tanA
Now e can see that
1+sec4A = tan4A/tan2A
1+sec8A = tan8A/tan4A

and the result drops right out.

Answer 2

how this happen tan(x/2)=sinx/(1+cosx)