Упростите выражение 4,5,6 заранее спасибо ))

[latex]1)\; \; (ctgx\cdot ctgy+1)\cdot cos(x+y)+(1-ctgx\cdot ctgy)\cdot cos(x-y)=\\\\=\frac{cosx\cdot cosy+sinx\cdot siny}{sinx\cdot siny}\cdot cos(x+y)+ \frac{sinx\cdot siny-cosx\cdot cosy}{sinx\cdot siny}\cdot cos(x-y)=\\\\= \frac{cos(x-y)\cdot cos(x+y)}{sinx\cdot siny}-\frac{cos(x+y)\cdot cos(x-y)}{sinx\cdot siny} =0[/latex] [latex]2)\; \; \frac{sin^2(x-y)+sin^2(x+y)}{2cos^2x\cdot cos^2y}-tg^2y=\\\\=\frac{(sinx\cdot cosy-siny\cdot cosx)^2+(sinx\cdot cosy+siny\cdot cosx)^2}{2cos^2x\cdot cos^2y} -tg^2y=\\\\= \frac{2sin^2x\cdot cos^2y+2sin^2y\cdot cos^2x}{2cos^2x\cdot cos^2y} -tg^2y=\\\\= \frac{2sin^2x\cdot cos^2y}{2cos^2x\cdot cos^2y} +\frac{2sin^2y\cdot cos^2x}{2cos^2x\cdot cos^2y} -tg^2y=tg^2x+tg^2y-tg^2y=tg^2x[/latex] [latex]3)\; \; ctg^2x\cdot ctg^2y- \frac{cos^2(x-y)+cos^2(x+y)}{2sin^2x\cdot sin^2y} =\\\\=ctg^2x\cdot ctg^2y- \frac{(cosx\cdot cosy+sinx\cdot siny)^2+(cosx\cdot cosy-sinx\cdot siny)^2}{2sin^2x\cdot sin^2y} =\\\\=ctg^2x\cdot ctg^2y- \frac{2cos^2x\cdot cos^2y+2sin^2x\cdot sin^2y}{2sin^2x\cdot sin^2y} =\\\\=ctg^2x\cdot ctg^2y-ctg^2x\cdot ctg^2y-1=-1[/latex]