Помогите решить, пожалуйста! срочно нужно!

Решите задачу:

$\frac{cos^2x-cos^2xsin^2x}{sin^2x} * \frac{cos^2x}{sin^2x-sin^2xcos^2x} = \frac{cos^2x(1-sin^2x)}{sin^2x} * \frac{cos^2x}{sin^2x(1-cos^2x)}$ $= \frac{cos^2x*cos^2x}{sin^2x} * \frac{cos^2x}{sin^2x*sin*2x} = \frac{cos^6x}{sin^6x} = ctg^6x$

$\left \{ {{ \frac{(x+y)(x+y)+(x-y)(x-y)}{x^2-y^2}= \frac{5}{2} } \atop {x^2+y^2=20}} \right.$
$\left \{ \frac{x^2+xy+xy+y^2+x^2-xy-xy+y^2}{x^2-y^2} \atop {x^2=20-y^2}} \right.$

$\left \{ {{2x^2+2y^2 = 2,5(x^2-y^2)} \atop {x^2=20-y^2}} \right.$
$\left \{ {{-0,5x^2+4,5y^2=0} \atop {x^2=20-y^2}} \right.$
$\left \{ {{10-0,5y^2 +4,5y^2=0} \atop {x^2=20-y^2}} \right.$
$\left \{ {{4y^2=10} \atop {x^2=20-y^2}} \right.$

$\left \{ {{y= +- \frac{ \sqrt{10} }{2} } \atop {x = +- \frac{ \sqrt{70} }{2} }} \right.$