If $$1 + 2\tan^{2}\theta+2\sin\theta\sec^{2}\theta=\frac{a}{b},0^\circ < \theta < 90^\circ$$, then $$\frac{a+b}{a-b}=$$?

A

$$\sin\theta$$

B

$$\cos\theta$$

C

$$\csc\theta$$

D

$$\sec$$