Question

$cos^4x+sin^4x+sin^2xcos^2x= cosxtgx/sin^2x - ctgxcosx= sinx*cosx*(tgx+ctgx)= (tgx+ctgx)^2-(tgx-ctgx)^2=$

Answer 1

1)(1+cos2x)²/4+(1-cos2x)²/4+(1-cos2x)(1+cos2x)/4=
=(1+2cos2x+cos²2x+1-2cos2x+cos²2x+1-cos²2x)/4=(3+cos2x)/4
2)cosx*sinx/cosx*sin²x -cosx*cosx/sinx=1/sinx - cos²x/sinx=(1-cos²x)sinx=sin²x/sinx=sinx
3)sinx*cosx(sin²x-cos²x)/sinx*cosx=-cos2x
4)(sin²x+cos²x)²/sin²xcos²x  -  (sin²x-cos²x)²/sin²xcos²x=
=((sin²x+cos²x)²-(sin²x-cos²x)²)sin²xcos²x=
=(sin²x+cos²x-sin²x+cos²x)(sin²x+cos²x+sin²x-cos²x)/sin²xcos²x=
=2cos²x*2sin²x/sin²²x*cos²x=4