$$\tan \theta = \frac{3}{4} \Rightarrow \tan 2\theta + \sec 2\theta = $$

$$\tan\ 2\theta\ +\sec\ 2\theta\ $$
$$=\frac{\left(1+\sin2\theta\ \right)}{\cos\ 2\theta\ }$$
$$\frac{\left(1+2\sin\ \theta\ \cos\ \theta\ \right)}{\cos\ ^2\theta\ -\sin\ ^2\theta\ }$$

Dividing both numerator and denominator by $$\cos\ ^2\theta$$ , we get

$$\frac{\left(\frac{1}{\cos\ ^2\theta}+2\tan\ \theta\ \right)}{1-\tan\ \theta\ }=\frac{\left(1+\tan^2\theta\ +2\tan\ \theta\ \right)}{1-\tan^2\theta\ }$$

Substitute the value of tan from the given data, we get option D as answer.