What is tan(θ/2), when tan(θ) = 4/3 and when π < θ < 3π/2?

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

tan (Ф/2)=-2; when tan (Ф)=4/3

Any pictures or something i need more info