prove that:  sec​80 - 1/ sec40 - 1 = tan80 / tan2​0

Dear student
$$\begin{gathered} To prove: \frac{sec8\theta -1}{sec4\theta -1}=\frac{\tan 8\theta }{\tan 2\theta } \\ LHS=\frac{sec8\theta -1}{sec4\theta -1} \\ =\frac{{\displaystyle \frac{1}{\cos 8\theta }}-1}{{\displaystyle \frac{1}{\cos 4\theta }}-1} \\ =\frac{1-\cos 8\theta }{\cos 8\theta }\times \frac{\cos 4\theta }{1-\cos 4\theta } \\ =\frac{2{\sin }^2\theta }{\cos 8\theta }\times \frac{\cos 4\theta }{2{\sin }^{2}2\theta } \left[\begin{array}{c}\because 1-\cos 8\theta =2{\sin }^2\frac{8\theta }{2}=2{\sin }^{2}4\theta \\ and, 1-\cos 4\theta =2{\sin }^2\frac{4\theta }{2}=2{\sin }^{2}2\theta \end{array}\right] \\ =\frac{2\sin 4\theta \cos 4\theta }{\cos 8\theta }\times \frac{\sin 4\theta }{2{\sin }^{2}2\theta } \\ =\frac{2sin4\theta cos4\theta }{cos8\theta }\times \frac{2\sin 2\theta \cos \theta }{2si{n}^{2}2\theta } \\ =\frac{\sin 2\left(4\theta \right)}{\cos 8\theta }\times \frac{\cos 2\theta }{\sin 2\theta } \\ =\frac{\sin 8\theta }{\cos 8\theta }\times \frac{\cos 2\theta }{\sin 2\theta } \\ =\tan 8\theta cot2\theta \\ =\frac{\tan 8\theta }{\tan 2\theta } \\ =RHS \end{gathered}$$
please mention that you are taking 0 as theta.
Regards