First,
tan(θ) = sin(θ) / cos(θ)
and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)
Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.
Now recall the half-angle identities,
cos²(θ/2) = (1 + cos(θ)) / 2
sin²(θ/2) = (1 - cos(θ)) / 2
and taking the positive square roots, we have
cos(θ/2) = √[(1 + cos(θ)) / 2]
sin(θ/2) = √[(1 - cos(θ)) / 2]
Then
tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]
Notice how we don't need sin(θ) ?
Now, recall the Pythagorean identity:
cos²(θ) + sin²(θ) = 1
Dividing both sides by cos²(θ) gives
1 + tan²(θ) = 1/cos²(θ)
We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.
cos²(θ) = 1/(1 + tan²(θ))
cos(θ) = - 1/√[1 + tan²(θ)]
Plug in tan(θ) = - 12/5 and solve for cos(θ) :
cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13
Finally, solve for sin(θ/2) and tan(θ/2) :
sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)
tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2
