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

Answer: tan (Ф/2) = - 2; when tan (Ф) = 4/3