If y sin ϕ = x sin (2θ + ϕ),

Given:

y sin ϕ = x sin (2θ + ϕ)

$\Rightarrow \frac{y}{x}=\frac{\sin (2 \theta+\phi)}{\sin \phi}$

Applying componendo and dividendo :

$\Rightarrow \frac{y-x}{y+x}=\frac{\sin (2 \theta+\phi)-\sin \phi}{\sin (2 \theta+\phi)+\sin \phi}$

$\Rightarrow \frac{y-x}{y+x} \Rightarrow \frac{2 \sin \left(\frac{2 \theta+\phi-\phi}{2}\right) \cos \left(\frac{2 \theta+\phi+\phi}{2}\right)}{2 \sin \left(\frac{2 \theta+\phi+\phi}{2}\right) \cos \left(\frac{2 \theta+\phi-\phi}{2}\right)}$

$\Rightarrow \frac{y-x}{y+x}=\frac{2 \sin \theta \cos (\theta+\phi)}{2 \sin (\theta+\phi) \cos \theta}$

$\Rightarrow \frac{y-x}{y+x}=\frac{\sin \theta \cos (\theta+\phi)}{\sin (\theta+\phi) \cos \theta}$

$\Rightarrow \frac{y-x}{y+x}=\frac{\cot (\theta+\phi)}{\cot \theta}$

$\Rightarrow(y-x) \cot \theta=(y+x) \cot (\theta+\phi)$