Question

tg^2a+ctg^2a-6/tg^2a+ctg^2a+2

Answer 1

Tg^2a+ctg^2a-6=sin²a/cos²a+cos²a/sin²a -6=
=(1-cos2a)/(1+cos2a)  +(1+cos2a)/(1-cos2a) - 6=
=(1-2cos2a+cos²2a+1+2cos2a+cos²2a -6+6cos²2a)/(1-cos²2a)=
=(8cos²2a-4)/(1-cos²2a)=4(2cos²2a-1)/sin²2a=4cos4a/sin²2a

Tg^2a+ctg^2a + 2=sin²a/cos²a+cos²a/sin²a +2=
=(1-cos2a)/(1+cos2a)  +(1+cos2a)/(1-cos2a) + 2=
=(1-2cos2a+cos²2a+1+2cos2a+cos²2a + 2- 2cos²2a)/(1-cos²2a)=
=4/(1-cos²2a)=4/sin²2a

(Tg^2a+ctg^2a - 6)/(Tg^2a+ctg^2a + 2)=4cos4a/sin²a:4/sin²a=
=4cos4a/sin²2a * sin²2a/4=cos4a