This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, please help

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

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