Log3 (2x+4)-log3 (2) = log3 (5)
log3 (2x+4)/2=log3 (5)
log3 (x+2) = log3 (5)
x+2=5, x=3
Кароче говоря, ответ - б.
(tg²α-sin²α)/(cos²α-ctg²α)=(sin²α/cos²α-sin²α)/(cos²α-cos²α/sin²α)=
=((sin²α*-sin²α*cos²α)/cos²α)/((sin²α*cos²α-cos²α)/sin²α)=
=(sin²α*(1-cos²α))/cos²α)/(cos²α*(sin²α-1))/sin²α)=(sin⁴α/cos²α)/(cos⁴α/sin²α)=
=sin⁶α/cos⁶α=tg⁶α.