Here is an idea of solution that I would like your comment on. Let's first write the following:
\begin{eqnarray}
\hspace{0.0cm}
A & = & \cos (\alpha) (1-\lambda) \\
B & = & \sin (\alpha) (1 + \lambda) \\
\cos (\alpha - \beta n ) - \lambda \, \cos (\alpha + \beta n ) & = & A \cos (\beta n) + B \sin ( \beta n ) \hspace{1.cm} \\
& = & \sqrt{A^2 + B^2} \, \sin ( a_0 + \beta n )
\end{eqnarray}
Where $a_0 \in (0, \frac{\pi}{2})$ ( since $\alpha \in (0, \frac{\pi}{2})$ ) .
\begin{eqnarray}
\hspace{0.0cm}
\sin ( a_0 ) & = & \frac{ A }{ \sqrt{A^2 + B^2} } \\
\cos ( a_0 ) & = & \frac{ B }{ \sqrt{A^2 + B^2} }
\end{eqnarray}
And therefore
\begin{eqnarray}
\hspace{0.0cm}
u_n & = & \frac{ \sqrt{A^2 + B^2} \, \sin ( a_0 + \beta n ) } { \cos^n (\beta ) } \hspace{1.cm}
\end{eqnarray}
We have two cases:
Case 1: $\beta = \frac{\pi p - a_0 }{q}$ and $a_0 = \frac{r \pi}{s}$ where $(r,s) = 1$, $p, q, r, s \in \mathbb{N}$.
In this case we have for each $n \in \mathbb{N}$:
\begin{eqnarray}
\sin ( a_0 + q (n s + 1) \beta ) & = & (-1)^{(n s + 1)p + 1} \sin ( \pi n r ) = 0 \hspace{1.0cm}
\end{eqnarray}
And therefore for each $n \in \mathbb{N}$: \begin{eqnarray}
u_{ q (n s + 1) } & = & 0 \hspace{1.cm}
\end{eqnarray}
And hence a subsequence of $(u_n)$ that converges to zero.
Case 2: $\beta = \lim_{m \to +\infty } \frac{s_m p_m - r_m }{s_m q_m} \pi $ and $a_0 = \lim_{m \to +\infty } \frac{r_m }{s_m} \pi $ where $(r_m,s_m) = 1$, $p_m, q_m, r_m, s_m \in \mathbb{N}$.
Thanks to the density of $\mathbb{Q}$ in $\mathbb{R}$, we can find the sequences $(\frac{\pi p_m - a_0 }{q_m})$ and $(\frac{\pi r_m }{s_m})$ that converges respectively to $\beta$ and $a_0$.
Without loss of generality, we can assume that the sequence $(q_m)$ is increasing.
Intuitively, if it works in Case 1, it should work in Case 2 thanks to the density of $\mathbb{Q}$ in $\mathbb{R}$.
For each $n$, we can write that:
\begin{eqnarray}
\sin ( a_0 + n \beta ) & = & \lim_{m \to +\infty } \sin ( \frac{r_m }{s_m} \pi + n \frac{s_m p_m - r_m }{s_m q_m} \pi ) \hspace{1.0cm} \\
u_n & = & \lim_{m \to +\infty } \frac{ \sqrt{A^2 + B^2} \, \sin ( \frac{r_m }{s_m} \pi + n \frac{s_m p_m - r_m }{s_m q_m} \pi ) } { \cos^n (\beta ) } \hspace{1.cm}
\end{eqnarray}
And
\begin{eqnarray} \hspace{-1.cm}
\lim_{n \to +\infty } u_n & = & \lim_{n \to +\infty } \lim_{m \to +\infty } \frac{ \sqrt{A^2 + B^2} \, \sin ( \frac{r_m }{s_m} \pi + n \frac{s_m p_m - r_m }{s_m q_m} \pi ) } { \cos^n (\beta ) } \hspace{1.cm}
\end{eqnarray}
If we take $n$ of the form $n_m = q_m ( 1 + m \prod^{m}_{i=1} s_i )$, we will have (subject to justification!):
\begin{eqnarray} \hspace{-1.cm}
\lim_{m \to +\infty } \sin ( a_0 + n_m \beta ) & = & \lim_{m \to +\infty } \underbrace{ \sin ( \frac{r_m }{s_m} \pi + q_m ( 1 + m \prod^{m}_{i=1} s_i ) \frac{s_m p_m - r_m }{s_m q_m} \pi ) }_{ = 0} \hspace{1.0cm} \\
& = & \lim_{m \to +\infty } 0 = 0
\end{eqnarray} And \begin{eqnarray} \hspace{-1.12cm}
\lim_{m \to +\infty } u_{ n_m } & = & \lim_{m \to +\infty } \underbrace{ \frac{ \sqrt{A^2 + B^2} \, \sin ( \frac{r_m }{s_m} \pi + q_m ( 1 + m \prod^{m}_{i=1} s_i ) \frac{s_m p_m - r_m }{s_m q_m} \pi ) } { \cos^{n_m} (\beta ) } }_{ = 0} \hspace{1.cm} \\
& = & \lim_{m \to +\infty } 0 = 0
\end{eqnarray}
Note that the sequence $(n_m)$ is strictly increasing since the sequence $(q_m)$ is increasing and the sequence $ ( 1 + m \prod^{m}_{i=1} s_i )$ is strictly increasing.
And hence a subsequence of $(u_n)$ that converges to zero.
Please let me know if there is any problem with such solution. Thank you.
