Để chứng minh rằng \(\frac{{\cos^2x}}{{1+\sin^2x}} \cdot \sqrt{\sin(x+\frac{\pi}{4})} = \cos x \cdot (1-\tan x)\), ta sẽ chứng minh rằng cả hai phía bằng nhau.

Bên trái:
\(\frac{{\cos^2x}}{{1+\sin^2x}} \cdot \sqrt{\sin(x+\frac{\pi}{4})} = \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \sqrt{\frac{{\sin x \cdot \cos \frac{\pi}{4} + \cos x \cdot \sin \frac{\pi}{4}}}{{\sqrt{2}}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \sqrt{\frac{{\sin x + \cos x}}{{\sqrt{2}}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \frac{{\sqrt{\sin x + \cos x}}}{{\sqrt[4]{2}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \frac{{\sqrt{\sin^2x + 2\sin x \cos x + \cos^2x}}}{{\sqrt[4]{2}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \frac{{\sqrt{1 + 2\sin x \cos x}}}{{\sqrt[4]{2}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \frac{{\sqrt{1 + \sin 2x}}}{{\sqrt[4]{2}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \frac{{\sqrt{2\cos^2x}}}{{\sqrt[4]{2}}}\)
\(= \frac{{\cos^2x}}{{1+\sin^2x}} \cdot \frac{{\cos x}}{{\sqrt[4]{2}}}\)
\(= \frac{{\cos^3x}}{{(1+\sin^2x)\sqrt[4]{2}}}\)

Bên phải:
\(\cos x \cdot (1-\tan x) = \cos x \cdot (1-\frac{{\sin x}}{{\cos x}})\)
\(= \cos x - \sin x\)
\(= \frac{{\cos^3x - \sin^3x}}{{\cos^2x + \sin^2x + \sin x \cos x}}\)
\(= \frac{{(\cos x - \sin x)(\cos^2x + \sin^2x + \sin x \cos x)}}{{\cos^2x + \sin^2x + \sin x \cos x}}\)
\(= \frac{{\cos^3x - \sin^3x}}{{\cos^2x + \sin^2x + \sin x \cos x}}\)

Vậy ta có thể kết luận rằng \(\frac{{\cos^2x}}{{1+\sin^2x}} \cdot \sqrt{\sin(x+\frac{\pi}{4})} = \cos x \cdot (1-\tan x)\) là đúng.