If $$\sec\theta+\tan\theta=p,\left(p>1\right)$$ then $$\frac{\operatorname{cosec}\theta+1}{\operatorname{cosec}\theta-1} = ?$$

$$\sec\theta+\tan\theta=p$$ ----(1)
$$\sec\theta-\tan\theta = \frac{1}{p}$$ ----(2)
From eq (1) and (2),
$$2\sec\theta = p + \frac{1}{p}$$
$$\sec\theta = \frac{p^2 + 1}{2p}$$ = $$\frac{hypotenuse}{base}$$
By the Pythagoras theorem,
$${hypotenuse}^2 = {base}^2 + {perpendicular}^2$$
$${p^2 + 1}^2 = {2p}^2 + {perpendicular}^2$$
$$perpendicular = \sqrt{p^4 + 1 - 2p^2}$$
$$perpendicular = {p^2 - 1}^2$$
Now,
$$\frac{\operatorname{cosec}\theta+1}{\operatorname{cosec}\theta-1}$$
$$\frac{\frac{p^2 + 1}{p^2 - 1}+1}{\frac{p^2 + 1}{p^2 - 1}-1}$$
$$\frac{p^2 + 1 + p^2 - 1}{p^2 + 1 - p^2 + 1}$$ = $$p^2$$