$$(1-\tan x)\sec{2x} \rightarrow \frac{1-\tan x}{\cos{2x}} \rightarrow \frac{1-\tan x}{\cos^2{x}-\sin^2{x}}\rightarrow \frac{1-\tan x}{(1-\tan^2{x})\cos^2{x}}\\\\\rightarrow \frac{1-\tan x}{(1-\tan{x})(1+\tan{x})\cos^2{x}}\rightarrow \frac{1}{(1+\tan{x})\cos^2{x}}$$
tan(pi/4) = 1 and cos(pi/4) = 1/√2 so:
$$\frac{1}{(1+\tan{\frac{\pi}{4}})\cos^2{\frac{\pi}{4}}}=\frac{1}{2\times \frac{1}{2}}=1$$
.
.$$(1-\tan x)\sec{2x} \rightarrow \frac{1-\tan x}{\cos{2x}} \rightarrow \frac{1-\tan x}{\cos^2{x}-\sin^2{x}}\rightarrow \frac{1-\tan x}{(1-\tan^2{x})\cos^2{x}}\\\\\rightarrow \frac{1-\tan x}{(1-\tan{x})(1+\tan{x})\cos^2{x}}\rightarrow \frac{1}{(1+\tan{x})\cos^2{x}}$$
tan(pi/4) = 1 and cos(pi/4) = 1/√2 so:
$$\frac{1}{(1+\tan{\frac{\pi}{4}})\cos^2{\frac{\pi}{4}}}=\frac{1}{2\times \frac{1}{2}}=1$$
.