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