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

We begin with the given integral

$$I=\int\frac{\tan x+\tan\alpha}{\tan x-\tan\alpha}\,dx,\qquad 0\lt \alpha\lt \dfrac{\pi}{2}\;( \alpha\ \hbox{constant}).$$

First we rewrite the numerator and the denominator with the help of the standard formula

$$\tan u\pm\tan v=\frac{\sin(u\pm v)}{\cos u\cos v}.$$

Applying this to $$u=x,\;v=\alpha$$, we obtain

$$\tan x+\tan\alpha=\frac{\sin(x+\alpha)}{\cos x\cos\alpha},\qquad \tan x-\tan\alpha=\frac{\sin(x-\alpha)}{\cos x\cos\alpha}.$$

Hence the entire fraction simplifies beautifully to

$$\frac{\tan x+\tan\alpha}{\tan x-\tan\alpha} =\frac{\dfrac{\sin(x+\alpha)}{\cos x\cos\alpha}} {\dfrac{\sin(x-\alpha)}{\cos x\cos\alpha}} =\frac{\sin(x+\alpha)}{\sin(x-\alpha)}.$$

Now we notice that the numerator can be written as a sine of a sum:

$$\sin(x+\alpha)=\sin\bigl[(x-\alpha)+2\alpha\bigr] =\sin(x-\alpha)\cos2\alpha +\cos(x-\alpha)\sin2\alpha.$$ This is the standard sine-addition identity $$\sin(A+B)=\sin A\cos B+\cos A\sin B.$$

Dividing the last expression by $$\sin(x-\alpha)$$ gives

$$\frac{\sin(x+\alpha)}{\sin(x-\alpha)} =\cos2\alpha+\sin2\alpha\, \frac{\cos(x-\alpha)}{\sin(x-\alpha)} =\cos2\alpha+\sin2\alpha\,\cot(x-\alpha).$$

Thus the integrand is split into two far simpler pieces:

$$\frac{\tan x+\tan\alpha}{\tan x-\tan\alpha} =\cos2\alpha+\sin2\alpha\cot(x-\alpha).$$

We can now integrate term by term. The first integral is trivial, while for the second we recall the basic antiderivative

$$\int\cot u\,du=\ln|\sin u|.$$

Since $$u=x-\alpha$$, we have $$du=dx$$, so

$$\begin{aligned} I &=\int\bigl[\cos2\alpha+\sin2\alpha\cot(x-\alpha)\bigr]\,dx \\ &=\cos2\alpha\int dx+\sin2\alpha\int\cot(x-\alpha)\,dx \\ &=\cos2\alpha\;x+\sin2\alpha\;\ln|\sin(x-\alpha)|+C. \end{aligned}$$

The constant $$C$$ absorbs every term that is independent of $$x$$, so subtracting (or adding) the fixed quantity $$\alpha\cos2\alpha$$ merely changes $$C$$. Therefore we may rewrite

$$I=\bigl(x-\alpha\bigr)\cos2\alpha+\sin2\alpha\;\ln|\sin(x-\alpha)|+C,$$ which is of the required form $$I=A(x)\cos2\alpha+B(x)\sin2\alpha+C.$$

By direct comparison we read off

$$A(x)=x-\alpha,\qquad B(x)=\ln|\sin(x-\alpha)|.$$

Hence, the correct answer is Option A.