Помогите решить логарифм и тождество...

Решите задачу:

$\log_2(3-x)+\log_2(1-x)=3\\O.D.3.:\\\begin{cases}3-x\ \textgreater \ 0\\1-x\ \textgreater \ 0\end{cases}\Rightarrow\begin{cases}x\ \textless \ 3\\x\ \textless \ 1\end{cases}\Rightarrow x\ \textless \ 1\\\log_2(3-x)(1-x)=3\\(3-x)(1-x) = 2^3\\3-4x+x^2=8\\x^2-4x-5=0\\D=16+4\cdot5=36\\x_{1,2}=\frac{4\pm6}2\\x_1=-1\\x_2=5\;-\;He\;nogx.\;no\;O.D.3.$

$tg\alpha+tg\beta=\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}=\frac{\sin\alpha\cos\beta+\sin\beta\cos\alpha}{\cos\alpha\cos\beta}=\\=\frac{\frac{\sin(\alpha-\beta)+\sin(\alpha+\beta)}2+\frac{\sin(\beta-\alpha)+\sin(\beta+\alpha)}2}{\cos\alpha\cos\beta}=\\=\frac{\sin(\alpha-\beta)+\sin(\alpha+\beta)-\sin(\alpha-\beta)+\sin(\alpha+\beta)}{2\cos\alpha\cos\beta}=\frac{2\sin(\alpha+\beta)}{2\cos\alpha\cos\beta}=\frac{\sin(\alpha+\beta)}{2\cos\alpha\cos\beta}$