If secθ + tanθ = p, (p > 1) then $$\frac{{{\text{cosec}}\,\theta + 1}}{{{\text{cosec}}\,\theta - 1}} = ?$$

$$\eqalign{ & \sec \theta + \tan \theta = p \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{p}{{\frac{1}{p}}} \cr & \frac{{\sec \theta + \tan \theta }}{{\sec \theta - \tan \theta }} = \frac{{{p^2}}}{1} \cr & {\text{Apply componendo and dividendo}} \cr & \frac{{\sec \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{\sec \theta .\cos \theta }}{{\tan \theta }} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & \frac{{{\text{cosec }}\theta }}{1} = \frac{{{p^2} + 1}}{{{p^2} - 1}} \cr & {\text{Apply again componendo and dividendo}} \cr & \frac{{{\text{cosec }}\theta + 1}}{{{\text{cosec }}\theta - 1}} = \frac{{{p^2}}}{1} \cr} $$