Tan (Ф/2)=⁺₋√[(1-cosФ)/(1+cosФ)]
if π<Ф<3π/2;
then, Where is Ф/2??
π/2<Ф/2<3π/4; therefore Ф/2 is in the second quadrant; then tan (Ф/2) will have a negative value.
tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
Now, we have to find the value of cos Ф.
tan (Ф)=4/3
1+tan²Ф=sec²Ф
1+(4/3)²=sec²Ф
sec²Ф=1+16/9
sec²Ф=(9+16)/9
sec²Ф=25/9
sec Ф=-√(25/9) (sec²Ф will have a negative value, because Ф is in the sec Ф=-5/3 third quadrant).
cos Ф=1/sec Ф
cos Ф=1/(-5/3)
cos Ф=-3/5
Therefore:
tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
tan(Ф/2)=-√[(1+3/5)/(1-3/5)]
tan(Ф/2)=-√[(8/5)/(2/5)]
tan(Ф/2)=-√4
tan(Ф/2)=-2
Answer: tan (Ф/2)=-2; when tan (Ф)=4/3
The third quartile of this data set is B. 24.
Answer:
y =-14
Step-by-step explanation:
Inverse variation is of the form
xy = k where k is a constant
(-7)(-8) = k
56 = k
The equation is
xy = 56
Let x = -4
(-4)y = 56
y = 56/-4
y =-14
