Sign error in relativistic velocity addition for angle transformation

[LCSphysicist]

Homework Statement

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Relevant Equations

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Adopt the speed of light equals one.
Calls $cos = c$, $sin = s$

$$ux' = \frac{v-uc}{1-uvc}$$
$$uy' = \frac{us}{\gamma(1-uvc)}$$
$$tan \theta' = uy' / ux' = \frac{us}{\gamma(v-uc)}$$

So that's basically my solution. The problem is: The answer is $\frac{us}{\gamma(v+uc)}$. Now, i can't understand why there is a plus sign instead my minus sign. Seems that, to got the answer provided, it was assumed that, for example, $ux' = \frac{v+uc}{1+uvc}$. Certainly wrong, since if v = uc, ux' should be zero.

So my question is, maybe my answer is right, the problem is that i assumed v to the right and the author assumed v to the left? Or did i a mistake? WHere?

 

[Orodruin]

Please show your work, not just the final result. It is impossible to help you and tell you where you have gone wrong if you do not provide this.

 

[LCSphysicist]

Please show your work, not just the final result. It is impossible to help you and tell you where you have gone wrong if you do not provide this.

Hello. The work is already showed, i am not sure what do you mean. I am just using the transformation of velocities in SR. Assuming that the velocity of particle u is $$\vec{u} = u cos (\theta) \hat{i} + u sin (\theta) \hat{j}$$ and $$\vec{v} = v \hat{i}$$

 

[PeroK]

I would say both are wrong. If $\theta = 0$ we have:$$u' = \frac{u - v}{1 - uv}$$