$\dfrac{1 - sin^{2}x}{2cot\left ( \dfrac{\pi}{4} + x \right ).cos^{2}\left ( \dfrac{\pi}{4} - x \right )} = 1$
[tex]\dfrac{1 - sin^{2}x}{2cot\left ( \dfrac{\pi}{4} + x \right ).cos^{2}\left ( \dfrac{\pi}{4} - x \right )} = \frac{cos{2x}}{2\frac{cos(\frac{\pi}{4}+x)}{sin(\frac{\pi}{4}+x)}.sin^{2}(x+\frac{\pi}{4})}=\frac{cos2x}{2.cos(x+\frac{\pi}{4})sin(x+\frac{\pi}{4})}=\frac{cos2x}{sin(2x+\frac{\pi}{2})}=1[/tex]