ПОМОГИТЕ..нужно упростить

Решите задачу:

$\frac{2cos^22a-1}{ctg(\frac{\pi}{4}-2a)\cdot cos^2(\frac{\pi}{4}+2a)} = \frac{cos4a}{ctg(\frac{\pi}{2}-(\frac{\pi}{4}+2a))cos^2(\frac{\pi}{4}+2a)}=\\\\=\frac{cos4a}{tg(\frac{\pi}{4}+2a)\cdot cos^2(\frac{\pi}{4}+2a)} = \frac{cos4a}{sin(\frac{\pi}{4}+2a)\cdot cos(\frac{\pi}{4}+2a)} =\\\\= \frac{cos4a}{\frac{1}{2}sin(\frac{\pi}{2}+4a)}=\frac{2cos4a}{cos4a} =2\\\\P.S.\; \; \; \; cos^2x= \frac{1+cos2x}{2} \; \; \to \; \; \; 2cos^2x-1=cos2x\; \; (x=2a\; ,\; \; 2x=4a)$