(i)[2]
By differentiating $\frac{1}{\cos x}$, demonstrate that when $y = \sec x$ then $\frac{dy}{dx} = \sec x \tan x$.
(ii)[1]
Show that $\dfrac{1}{\sec x - \tan x} = \sec x + \tan x$.
(iii)[2]
Deduce that $\dfrac{1}{(\sec x - \tan x)^2} = 2\sec^2 x - 1 + 2\sec x \tan x$.
(iv)[3]
Hence show that $\int_0^{\frac{1}{4}\pi} \frac{1}{(\sec x - \tan x)^2}\,dx = \frac{1}{4}(8\sqrt{2} - \pi)$.