\(\frac{2cos^2x-1}{2cot\left(\pi+\frac{\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}=\frac{cos2x}{2cot\left(\frac{\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}=\frac{cos2x}{2cos\left(\frac{\pi}{4}+x\right).sin\left(\frac{\pi}{4}+x\right)}\)
\(=\frac{cos2x}{sin\left(\frac{\pi}{2}+2x\right)}=\frac{cos2x}{cos2x}=1\)