If secθ + tanθ = 5, then the value of the $$\frac{\tan\theta+1}{\tan\theta-1}$$

sec$$^2 \theta $$ - tan$$^2 \theta$$ =1
(sec$$\theta$$ + tan$$\theta$$)(sec$$\theta$$ - tan $$\theta$$) = 1
sec$$\theta$$ - tan$$\theta$$ = $$\frac{1}{5}$$--------------1
sec$$\theta$$ + tan$$\theta$$ = $$5$$ -------------------2
Adding 1 and 2
2sec$$\theta$$ = $$ 5+ \frac{1}{5}$$
sec$$\theta$$ = $$\frac{26}{10}$$
Hence tan$$\theta$$ = $$\frac{24}{10}$$
tan$$\theta$$ + 1 = $$\frac{24}{10} + 1 = \frac{34}{10}$$
tan$$\theta$$ - 1 = $$\frac{24}{10} - 1 = \frac{14}{10}$$

$$\frac{tan \theta +1}{tan \theta -1} = \frac{ \frac{34}{10} } { \frac{14}{10} } $$

$$ = \frac{34}{14}$$
$$ = \frac{17}{7}$$
Option D is the correct answer.