Помогите пожалуйста решить 1 и 2 задания

1сtg²a*ctg²b-(cos(a+b)*cos(a-b))/(sin²a*sin²b)==ctg²a*ctg²b-[(cosacosb-sinasinb)(cosacosb+sinasinb)/(sin²asin²b)]==ctg²a*ctg²b-(cos²acos²b-sin²asin²b)/(sin²a*sin²b)==ctg²a*ctg²b-cos²a*cos²b/(sin²a*sin²b)+sin²a*sin²b/(sin²a*sin²b)==ctg²²a*ctg²b-ctg²a*ctg²b+1=12ОДЗx-2>0⇒x>2x-4≠0⇒x≠0x∈(2;4) U (4;∞)2log(5)(x-2)=-log(5)([(x-4)²]2log(5)(x-2)=-2log(5)(x-4)2log(5)(x-2)+2log(5)(x-4)=02[log(5)(x-2)+log(5)(x-4)]=0log(5)[(x-2)(x-4)]=0(x-2)(x-4)=1x²-4x-2x+8=1x²-6x+7=0D=36-28=8x1=(6-2√7)/2=3-√7∉ОДЗx2=3+√7