Given $\tan \alpha=\frac{4}{3}$ and $\int \frac{1}{3 \cos x-4 \sin x} d x$
$$
\begin{aligned}
& I=\int \frac{d x}{3 \cos x-4 \sin x} \\
& I=\frac{1}{3} \int \frac{d x}{\cos x-\frac{4}{3} \sin x}=\frac{1}{3} \int \frac{d x}{\cos x-\tan \alpha \sin x} \\
& I=\frac{1}{3} \int \frac{\cos \alpha d x}{\cos \alpha \cos x-\sin \alpha \sin x} \\
& I=\frac{1}{3} \int \frac{\cos \alpha d x}{\cos (x+\alpha)} \\
& I=\frac{\cos x}{3} \int \sec (x+\alpha) d x
\end{aligned}
$$
Here, $\cos \alpha=\frac{3}{5}$
$$
\begin{aligned}
& I=\frac{3}{5 \times 3} \times \log |\sec (x+\alpha)+\tan (x+\alpha)|+c \\
& I=\frac{1}{5} \log \left|\frac{1+\sin (x+\alpha)}{\cos (x+\alpha)}\right|+c \\
& I=\frac{1}{5} \log \left|\frac{\sin ^2 \frac{(x+\alpha)}{2}+\cos ^2 \frac{(x+\alpha)}{2}+2 \sin \frac{(x+\alpha)}{2} \cos \frac{(x+\alpha)}{2}}{\cos ^2 \frac{(x+\alpha)}{2}-\sin ^2 \frac{(x+\alpha)}{2}}\right| \\
& I=\frac{1}{5} \log \left|\frac{\left.\sin \frac{(x+\alpha)}{2}+\cos \frac{(x+\alpha)}{2}\right)}{\cos \frac{(x+\alpha)}{2}-\sin \frac{(x+\alpha)}{2}}\right|
\end{aligned}
$$
$$
I=\frac{1}{5} \log \left|\frac{1+\tan \frac{(x+\alpha)}{2}}{1-\tan \frac{(x+\alpha)}{2}}\right|=\frac{1}{5} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{\alpha}{2}\right)\right|+c
$$
Option (b) is correct.