Consider the following:
\begin{align*}
c-a\tan \theta & = b \sec \theta\\
(c-a\tan \theta)^2 & = (b\sec \theta)^2\\
(a^2-b^2)\tan^2 \theta -2ac \tan \theta+(c^2-b^2) & = 0.
\end{align*}
Think of this as a quadratic in $\tan \theta$. It has two solutions $\tan \alpha$ and $\tan \beta$. So
\begin{align*}
\tan \alpha + \tan \beta & = \frac{2ac}{a^2-b^2}\\
\tan \alpha \cdot \tan \beta & = \frac{c^2-b^2}{a^2-b^2}
\end{align*}
Now you can compute using
$$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$
The equation of the line $PQ$ is given by
$$y-a\tan\alpha=\frac{a\tan\alpha-a\tan\beta}{a\sec\alpha-a\sec\beta}(x-a\sec\alpha),$$
i.e.
$$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}x+\frac{\cos\alpha-\cos\beta}{\sin\alpha-\sin\beta}y=a\tag3$$
The $(2)$ you wrote can be written as
$$\frac{\cos\alpha}{2}x+\frac{1}{2\tan\alpha}y=a\tag4$$
Comparing $(3)$ with $(4)$ gives
$$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}=\frac{\cos\alpha}{2},$$
i.e.
$$2\sin\alpha-\cos\alpha\tan\beta=\cos\alpha\sin\alpha\sec\beta\tag5$$
Squaring the both sides and using $\sec^2\beta=1+\tan^2\beta$ gives
$$(\cos^2\alpha\sin^2\alpha-\cos^2\alpha)\tan^2\beta+4\sin\alpha\cos\alpha\tan\beta+\cos^2\alpha\sin^2\alpha-4\sin^2\alpha=0$$
and so
$$\begin{align}&\tan\beta\\&=\frac{-2\sin\alpha\cos\alpha\pm\sqrt{4\sin^2\alpha\cos^2\alpha-(\cos^2\alpha\sin^2\alpha-\cos^2\alpha)(\cos^2\alpha\sin^2\alpha-4\sin^2\alpha)}}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha\pm\sqrt{\cos^2\alpha\sin^2\alpha(\cos^2\alpha-2)^2}}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha\pm \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha - \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha},\quad \frac{-2\sin\alpha\cos\alpha + \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\tan\alpha,\quad \tan\alpha(4\sec^2\alpha-1)\end{align}$$
Now when $\tan\beta=\tan\alpha$,
$$\begin{align}(5)&\Rightarrow \sin\alpha(\cos\beta-\cos\alpha)=0\\&\Rightarrow \sin\alpha=0\quad\text{or}\quad \cos\beta=\cos\alpha\end{align}$$
Case 1 : $\sin\alpha=0\Rightarrow \tan\alpha=\tan\beta=0\Rightarrow (\alpha,\beta)=(0,\pi),(\pi,0)\qquad (\text{$\tan\beta=\tan\alpha(4\sec^2\alpha-1)$ holds in this case})$
Case 2 : $\cos\beta=\cos\alpha\Rightarrow \sin\alpha=\sin\beta\Rightarrow \alpha=\beta\qquad (\text{this case is eliminated})$
Thus, we get that $\tan\beta=\tan\alpha(4\sec^2\alpha-1)$.
Best Answer
$$\frac{2}{\cos\alpha\cos\beta\sin\alpha\sin\beta} = \frac{8}{\sin2\alpha\sin2\beta} \geq 8$$
Now why did the previous inequality only give 2 whereas when we use this we get 8? Basically, $\alpha$ and $\beta$ are independent of each other, hence we can minimise the second expression by putting $\alpha = \beta = \frac{\pi}{4}$
On the other hand, in the expression you reduced to, both $\sin\alpha$ and $\cos\alpha$ cannot simultaneously be $1$, hence the product will actually have a different maximum value (which is $\frac{1}{2}$)