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Mathematics
Mathematics
Problem
Prove that
sec
8
θ
−
1
sec
4
θ
−
1
=
tan
8
θ
tan
2
θ
\frac{\sec 8\theta - 1}{\sec 4\theta - 1} = \frac{\tan 8\theta}{\tan 2\theta}
s
e
c
4
θ
−
1
s
e
c
8
θ
−
1
=
t
a
n
2
θ
t
a
n
8
θ
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