1. Find the root sign (1 + 2 * sin610 ° * cos430 °) / sin 250 ° + cos 790 ° PS: the root sign is 1 + 2 * sin610 ° * cos430 ° 2. Given Tan a = - 3, find the value of the following formula: (1)(4*sin a-2*cos a)/(5*cos a=3*sin a) (2) (sin a-cos a)^2 There should be a specific process

1. Find the root sign (1 + 2 * sin610 ° * cos430 °) / sin 250 ° + cos 790 ° PS: the root sign is 1 + 2 * sin610 ° * cos430 ° 2. Given Tan a = - 3, find the value of the following formula: (1)(4*sin a-2*cos a)/(5*cos a=3*sin a) (2) (sin a-cos a)^2 There should be a specific process

1.√(1+2*sin610°*cos430°)/sin 250°+cos 790°
=√(1+2*sin(-110)°*cos70°)/sin110°+cos70°
=√(1-2*sin110°*cos70°)/sin110°+cos70°
Let 110 ° be x and 70 ° be y
2*sin110°*cos70°=2*sinx*cosy=2*(1/2)*[sin(x+y)+sin(x-y)]=sin180°+sin40°=sin40°=sin(2*20°)=2*sin20°cos20°
√(1+2*sin610°*cos430°)=√1+2*sin20°cos20°=√（cos20+sin20）^2
=cos20°+sin20°
cos20°+sin20°=sin70°+sin20°
sin110°+cos70°=sin70°+sin20°
So the original formula = (sin70 ° + sin20 °) / (sin70 ° + sin20 °) = 1
2.tana=-3
So Sina / cosa = - 3
(1)(4*sin a-2*cos a)/(5*cos a=3*sin a)
Then the original formula = (4 (Sina / COSA) - 2) / (5-3 (Sina / COSA))
=(4*(-3)-2)/(5-3(-3))
=-14/14
=-1
(sin a-cos a)^2
=(sina)^2+(cosa)^2-2sina*cosa
=1-2sina*cosa
=1-sin2a
=1-[(2tana)/(1+(tana)^2)]
=1-(-6/10)
=8/5