If $$0 < \theta < 90^\circ$$, $$3b\operatorname{cosec}\theta=a\sec\theta$$ and $$3a\sec\theta-b\operatorname{cosec}\theta=8$$, then the value of $$9b^2 + a^2$$ is:

Given,  $$3b\operatorname{cosec}\theta=a\sec\theta$$

$$\Rightarrow$$  $$b\operatorname{cosec}\theta=\frac{a\sec\theta}{3}$$

$$3a\sec\theta-b\operatorname{cosec}\theta=8$$

$$\Rightarrow$$  $$3a\sec\theta-\frac{a\sec\theta\ }{3}=8$$

$$\Rightarrow$$  $$a\sec\theta\left(3-\frac{1\ }{3}\right)=8$$

$$\Rightarrow$$  $$a\sec\theta\left(\frac{8}{3}\right)=8$$

$$\Rightarrow$$  $$a\sec\theta=3$$ .................(1)

$$3b\operatorname{cosec}\theta=a\sec\theta$$

$$\Rightarrow$$  $$\frac{3b}{\sin\theta\ }=\frac{a}{\cos\theta\ }$$

$$\Rightarrow$$  $$a\tan\theta\ =3b$$

$$\Rightarrow$$  $$a^2\tan^2\theta\ =9b^2$$

$$\Rightarrow$$  $$a^2\left(\sec^2\theta\ -1\right)=9b^2$$

$$\Rightarrow$$  $$a^2\sec^2\theta\ -a^2=9b^2$$

$$\Rightarrow$$  $$3^2-a^2=9b^2$$ (From 1)

$$\Rightarrow$$  $$9b^2+a^2=9$$

Hence, the correct answer is Option B