Ответ:

[tex]3) \ sin^{2}(x)-cos^{2}(x) = -(cos^{2}(x) - sin^{2}(x))=-cos(2x)\\\\6) \ \frac{cos(a)-sin(2a)}{1-2sin(a)}=\frac{cos(a)-2sin(a)cos(x)}{1-2sin(a)}=\frac{cos(a)(1-2sin(x))}{1-2sin(a)}=cos(x)\\\\9) \frac{2cos(x)cos(2x)}{ctg(2x)}=\frac{2cos(x)cos(2x)}{\frac{cos(2x)}{sin(2x)}}=2cos(x)cos(2x)\frac{sin(2x)}{cos(2x)}=\frac{2cos(x)cos(2x)sin(2x)}{cos(2x)}=2cos(x)sin(2x)[/tex]