We have $\cos (\theta+\phi)=m \cos (\theta-\phi)$ $\Rightarrow \frac{\cos (\theta+\phi)}{\cos (\theta-\phi)}=\frac{m}{1}$

Using componendo and dividendo,
$$
\Rightarrow \quad \frac{\cos (\theta-\phi)-\cos (\theta+\phi)}{\cos (\theta-\phi)+\cos (\theta+\phi)}=\frac{1-m}{1+m}
$$


$$
\Rightarrow \frac{-2 \sin \left(\frac{\theta-\phi+\theta+\phi}{2}\right) \cdot \sin \left(\frac{\theta-\phi-\theta-\phi}{2}\right)}{2 \cos \left(\frac{\theta-\phi+\theta+\phi}{2}\right) \cdot \cos \left(\frac{\theta-\phi-\theta-\phi}{2}\right)}=\frac{1-m}{1+m}
$$
$$
\Rightarrow \frac{\sin \theta \cdot \sin \phi}{\cos \theta \cdot \cos \phi}=\frac{1-m}{1+m} \quad\left[\begin{array}{l}
\because \sin (-\theta)=-\sin \theta \\
\text { and } \cos (-\theta)=\cos \theta
\end{array}\right]
$$
$$
\Rightarrow \tan \theta \cdot \tan \phi=\frac{1-m}{1+m}
$$
$\Rightarrow \tan \theta=\left(\frac{1-m}{1+m}\right) \cot \phi$
Hence proved.