A photon of wavelength equal to the Compton wavelength λC=h/(mec) of the electron is scattered by a free electron initially at rest, through an angle…

Compton shift: $\lambda'-\lambda=\lambda_C(1-\cos90^\circ)=\lambda_C$, so $\lambda'=2\lambda_C$. Photon momenta are $p_0=h/\lambda_C=m_ec$ (incident) and $p'=h/(2\lambda_C)=p_0/2$ (scattered, perpendicular to the incident direction). Conservation of momentum gives the electron momentum components: along the incident direction $p_x=p_0$, and perpendicular $p_y=-p'=-p_0/2$ (opposite to the scattered photon). Then $\tan\varphi=|p_y|/p_x=\tfrac12$, so $\cot\varphi=2$. Consistency check with energy: the electron gains $E_0-E'=m_ec^2-\tfrac12m_ec^2=\tfrac12 m_ec^2$, so $E_e=1.5\,m_ec^2$; and $p_e^2c^2=p_0^2c^2\,(1+\tfrac14)=1.25\,m_e^2c^4$, giving $E_e^2=m_e^2c^4+p_e^2c^2=2.25\,m_e^2c^4$, as required. The kinematics can also be seen from the vector triangle: the incident photon momentum $\vec p_0$ equals the sum of the scattered photon momentum $\vec p'$ and the electron momentum $\vec p_e$, so $\vec p_e=\vec p_0-\vec p'$ has magnitude $\sqrt{p_0^2+p'^2}=\tfrac{\sqrt5}{2}p_0$ and its direction is below the incident axis by $\arctan(1/2)\approx26.6^\circ$. Because the scattering angle is $90^\circ$ the two photon momenta are perpendicular and no trigonometric ambiguity arises.