It is known that a.b.c. is the three internal angles of a.b.c. the vector M = (- 2,1), n = (COS (a + π / 6), sin (a - π / 3)), and M is perpendicular to n Find the angle a if Sin & # 178; c-cos & # 178; C / (1-sin2c) = - 2, find the value of tanb

It is known that a.b.c. is the three internal angles of a.b.c. the vector M = (- 2,1), n = (COS (a + π / 6), sin (a - π / 3)), and M is perpendicular to n Find the angle a if Sin & # 178; c-cos & # 178; C / (1-sin2c) = - 2, find the value of tanb

one
m=(-2,1),n=(cos(A+π/6),sin(A-π/3))
M ⊥ n, i.e. m · n = (- 2,1) · (COS (a + π / 6), sin (a - π / 3))
=-2cos(A+π/6)+sin(A-π/3)
=-2(√3cosA/2-sinA/2)+(sinA/2-√3cosA/2)
=3sinA/2-3√3cosA/2
=3sin(A-π/3)=0
That is: sin (a - π / 3) = 0
A ∈ (0, π), that is: a - π / 3 ∈ (- π / 3,2 π / 3)
That is: a - π / 3 = 0
That is: a = π / 3
two
(sin²C-cos²C)/(1-sin2C)
=(sinC+cosC)(sinC-cosC)/(sinC-cosC)^2
=(sinC+cosC)/(sinC-cosC)=-2
That is: 3sinc = COSC
That is: Tanc = 1 / 3
B + C = 2 π / 3, that is, B = 2 π / 3-c
That is: tanb = Tan (2 π / 3-C)
=(tan(2π/3)-tanC)/(1+tan(2π/3)tanC)
=(-√3-1/3)/(1-√3/3)
=-(6+5√3)/3