2(sin^2)14/7∏+(tan^2)7/3∏cot(-7/4∏),

Question

2(sin^2)14/7∏+(tan^2)7/3∏cot(-7/4∏),

Answer 1

原式=2[sin^2(2∏)+tan^2(∏+∏+∏/3)*cot{[∏＋(3/4)∏]}
原式=0+tan^2(∏/3)*[-cot(3/4)〕
=tan^2(∏/3)*[-cot(∏-∏/4)]
=(根号3)^2*[-(-cot∏/4
=3*1
=3
答：原式=3