Determining the phase delay between H and E fields I want to determine the  phase delay between H and E fields in a medium with losses and not electrically charged. In this medium we also have $\sigma_c\approx \varepsilon \omega$. The enunciation of the problem gives the following hint:
$$\tan(2\phi)=\frac{2}{\text{cotan}{\phi}-\tan{\phi}}$$
The solution of this problem is:
$$\phi=\frac{1}{2}\arctan{\left(\frac{\sigma_c}{\varepsilon \omega}\right)}$$
I will show you my first steps on the resolution:

We have:
$$\vec{\bar{E}}=\bar{E_0}e^{\left(\bar{k}z-wt\right)} ,\hspace{15pt}\vec{\bar{H}}=\bar{H_0}e^{\left(\bar{k}z-wt\right)}$$
where the bar symbol is due to complex notation.
From Maxwell's equation on the reciprocal space:
$$\left[\vec{k}\times \vec{E}\right]=w\mu \vec{H}$$
So, by knowing that $\vec{E}, \vec{H}, \vec{k}$ are perpendicular to each other then we can say (after some computing):
$$\bar{H_0}=\frac{\bar{k}\bar{E_0}}{w\mu}$$
By complex notation we can define:
$$\bar{E_0}=E_0e^{i\delta_E}\hspace{2pt} \text{,  so}\hspace{3pt}\bar{H_0}=H_0e^{\left(i\delta_E+\phi\right)}$$
where $\phi$ is due to $\bar{k}$ phase ($\bar{k}=ke^{i\phi}$).
The phase delay between $H$ and $E$ will then be $\phi$. And we can calculate $\phi$ by the following way:
$$\phi=\arctan\left(\frac{k_i}{k_r}\right)$$
where $k_i=\omega\sqrt{\frac{\varepsilon \mu}{2} \left(-1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega} \right)^2}\right)}$ and $k_r=\omega\sqrt{\frac{\varepsilon \mu}{2} \left(1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega} \right)^2}\right)}$.

My question here is where should I use the approximation $\sigma_c\approx \varepsilon \omega$? Because if I use it directly I can make $\frac{\sigma_c}{\varepsilon \omega}=1$ which will not give me the solution that I enunciated above. 
I know that I need to use the hint. The only problem is what to do after the last step of my resolution.
 A: I just did it!
We have that $$\tan(2\phi)=\frac{2}{\text{cotan}{\phi}-\tan\phi}\Leftrightarrow$$
$$\Leftrightarrow\arctan\left[\tan(2\phi)\right]=\arctan\left\{\frac{2}{\text{cotan}\left[\arctan\left(\frac{ki}{kr}\right)\right]-\tan\left[\arctan\left(\frac{ki}{kr}\right)\right]}\right\}\Leftrightarrow$$
$$\Leftrightarrow \phi=\frac{1}{2}\arctan\left(\frac{2}{\frac{kr}{ki}-\frac{ki}{kr}}\right)$$
And,
$$\frac{2}{\frac{kr}{ki}-\frac{ki}{kr}}=2\frac{\sqrt{-1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega}\right)^2}}. \sqrt{1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon  \omega}\right)^2}}}{1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon  \omega}\right)^2}-\left(-1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon  \omega}\right)^2}\right)}=2\frac{\frac{\sigma_c}{\varepsilon\omega}}{2}=\frac{\sigma_c}{\varepsilon\omega}$$
So, $$\phi=\frac{1}{2}\arctan\left(\frac{\sigma_c}{\varepsilon\omega}\right)$$
which is the solution of the problem. We didn't even need the approximation $\sigma_c \approx \varepsilon \omega$.
