Question

2sin^(-2490)+tan 1410

Answer 1

2sin²(-2490)+tan 1410=2sin²(-330-6*360)+tan (330+3*360)=
=2sin²(-330)+tan (330)=2 sin²(30-1*360)+tan (-30)=
=2sin²(30)-tan (30)=2(1/2)²-√3/3=1/2-√3/3)=(3-2√3)/6

Answer: 2 sin²(-2490)+tan (1410)=(3-2√3)/6    (≈-0.07735...)