= Solution
Use <four-momentum> with components $(E/c,p_x,p_y)$, suppressing the identically zero third spatial component. The <photon> in the primed frame has
$$
\boxed{P'=\frac{h\nu'}c(1,\cos\theta',\sin\theta').}
$$
Its <energy> is $h\nu'$ and its spatial <momentum> has magnitude $h\nu'/c$. Apply the same <Lorentz transformation> as for the spacetime coordinates. With $\beta=v/c$,
$$
P=\frac{h\nu'}c
\left(\gamma(1+\beta\cos\theta'),\
\gamma(\cos\theta'+\beta),\
\sin\theta'\right).
$$
The time component gives the <relativistic Doppler effect>
$$
\boxed{\nu=\gamma\nu'(1+\beta\cos\theta').}
$$
The <aberration of light> follows by dividing spatial components by $E/c$:
$$
\boxed{\cos\theta=\frac{\cos\theta'+\beta}{1+\beta\cos\theta'},\qquad
\sin\theta=\frac{\sin\theta'}{\gamma(1+\beta\cos\theta')}.}
$$
In particular
$$
\tan\theta=\frac{\sin\theta'}{\gamma(\cos\theta'+\beta)}
=\frac{\tan\theta'}{\gamma(1+\beta\sec\theta')},
$$
where the tangent expression is understood only when its denominators exist. The sine and cosine equations determine the correct quadrant and also cover directions where the tangent/secant expression is undefined.
For the <photon-number half-cone under relativistic aberration>, the forward half of the primed emission angles is $-\pi/2\leq\theta'\leq\pi/2$. The <aberration of light> is monotone on this interval, and its endpoints have $\cos\theta=\beta$. Thus exactly half of a uniform angular <photon> distribution lies within the lab-frame cone
$$
\boxed{|\theta|\leq\alpha,\qquad \cos\alpha=\frac vc.}
$$
As $v\to c^-$, $\alpha=\arcsin(1/\gamma)\sim1/\gamma$, so the cone becomes narrow. Here “half” counts emitted photons or ray directions; it is not a claim that half of the observed <energy> or power lies there, since the <relativistic Doppler effect> changes the <photon> energies with angle.
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