Prove that. (1+sec2A) (1+sec4A) (1+sec8A)=Tan8A×CotA

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 we can see that
1+sec4A = tan4A/tan2A
1+sec8A = tan8A/tan4A

Now multiply and get the result