If secθ + tanθ = p, (p > 1) then cosec\, + 1 cosec\, - 1 = ?

Answer: Option C

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