If a = Tan (- 7 π / 6), B = cos23 π / 4, C = sin (- 33 π / 4), the comparison is large

If a = Tan (- 7 π / 6), B = cos23 π / 4, C = sin (- 33 π / 4), the comparison is large

tan(2π-7π/6)=tan(5π/6)=-√3/3
cos23π/4=cos(6π-π/4)=cos(-π/4)=cos(π/4)=√2/2
sin(-33π/4)=sin(-π/4)=-√2/2
Root 3 = 1.7, root 2 = 1.4
So cos23 π / 4 > Tan (2 π - 7 π / 6) > sin (- 33 π / 4)

How to compare the value of sin cos Tan in the same angle

0°

It is known that sin α + cos α = - 1 / 5, α ∈ (- TII / 2, TII / 2), then Tan α=
The answer is - 4 / 3, I calculate α in the second quadrant, sin α = 3 / 5, cos α = - 4 / 5, Tan α = - 3 / 4, what's wrong?

You can divide the whole formula by one, that is, sin α square + cos α square = 1, and the relationship between the numerator denominator and Tan α, that is, divided by cos α square, can be solved