It is known that the coordinates of three points a, B and C are a (3,0), B (0,3), C (COS α, sin α), (α ≠ K π 4, K ∈ z), respectively. If AC · BC = - 1, then the value of 1 + sin2 α − Cos2 α 1 + Tan α is 1______ ．

It is known that the coordinates of three points a, B and C are a (3,0), B (0,3), C (COS α, sin α), (α ≠ K π 4, K ∈ z), respectively. If AC · BC = - 1, then the value of 1 + sin2 α − Cos2 α 1 + Tan α is 1______ ．

AC = (COS α - 3, sin α), BC = (COS α, sin α - 3). ∵ AC · BC = - 1, ∵ cos α (COS α - 3) + sin α (sin α - 3) = - 1, changed to sin α + cos α = 23. ∵ 49 = (sin α + cos α) 2 = 1 + 2Sin α cos α, changed to 2Sin α cos α = - 59. ∵ 1 + sin2 α − Cos2 α 1 + ta

(6sin a + 5cos a) / (5cos a + 3sin a) = 8, then Tan a=__
Can this problem be reduced to 6sin (a) + 5cos (a) = 40cos (a) + 24sin (a)
-35cos(a)=18sin(a)
Then compare Tan a
If not, please tell me,

(6sin a+5cos a)/(5cos a+3sin a)
={(6sin a+5cos a)/sina}/{(5cos a+3sin a)/sina}
=[6+5ctga]/[5ctga+3]=8
CTGA = - 18 / 35
ctga=1/tga=-35/18

3sin θ - 4cos θ = - 5cos (θ + φ), find Tan φ

∵√(3^2+4^2)=5.
∵3sinθ-4cosθ=5[(3/5)sinθ-(4/5)cosθ].
=5(sinθcosφ-cosιθsinφ)
=5sin(θ-φ)
Where cos φ = 3 / 5 and sin φ = 4 / 5
The formula to the right of the equal sign of the original formula is redundant