Let \(\alpha \in\left(0, \frac{\pi}{2}\right)\) be fixed. If the integral \(\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} d x=A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+c\), (where \(\mathrm{c}\) is a constant of integration), then functions \(\mathrm{A}(x)\) and \(\mathrm{B}(x)\) are respectively

\(\text {Let } \mathrm{I} =\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} \mathrm{d} x \)
\( =\int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x}-\frac{\sin \alpha}{\cos \alpha}} \mathrm{d} x \)
\( =\int \frac{\sin x \cos \alpha+\sin \alpha \cos x}{\sin x \cos \alpha-\sin \alpha \cos x} \mathrm{~d} x \)
\( =\int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} \mathrm{d} x\)
Let \(x-\alpha=\mathrm{t}\)
\(\therefore\mathrm{I} =\frac{\sin (\mathrm{t}+2 \alpha)}{\sin \mathrm{t}} \)
\( =\int \frac{\sin (\mathrm{t}) \cos 2 \alpha+\cos (\mathrm{t}) \sin 2 \alpha}{\sin (\mathrm{t})} \mathrm{d} x \)
\( =\cos 2 \alpha \int 1 \mathrm{dt}+\sin 2 \alpha \int \cot (\mathrm{t}) \mathrm{dt} \)
\( =\cos 2 \alpha \cdot \mathrm{t}+\sin 2 \alpha \cdot \log |\sin (\mathrm{t})|+\mathrm{c} \)
\( \therefore \mathrm{I} =(x-\alpha) \cos 2 \alpha+\log |\sin (x-\alpha)| \sin 2 \alpha+\mathrm{c} \)
\( \mathrm{But} \int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} \mathrm{d} x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\) \(\mathrm{c} ... [Given] \)
\( \Rightarrow \mathrm{A}(x)=x-\alpha, \mathrm{B}(x)=\log |\sin (x-\alpha)|+\mathrm{c}\)