Prove that $\frac{\sin(x+\theta)}{\sin(x+\phi)} = \cos(\theta - \phi) +\cot(x+\phi)\sin(\theta-\phi)$. 
Prove that $$\frac{\sin(x+\theta)}{\sin(x+\phi)} = \cos(\theta - \phi) +\cot(x+\phi)\sin(\theta-\phi).$$

I tried applying componendo and dividendo on LHS but couldn't prove. Please help.
Thank you.
 A: $$\frac{\sin(x+\theta)}{\sin(x+\phi)}=\frac{\sin[(x+\phi)+(\theta-\phi)]}{\sin(x+\phi)}=\frac{\sin(x+\phi)\cos(\theta-\phi)+\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}=\\
=\frac{\sin(x+\phi)\cos(\theta-\phi)}{\sin(x+\phi)}+\frac{\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}=\cos(\theta-\phi)+\cot(x+\phi)\sin(\theta-\phi)$$
for the last equality I used $\cot(x+\phi)=\frac{\cos(x+\phi)}{\sin (x+\phi)}$
A: Multiply both sides by $\sin(x+ϕ)$:
$$\sin(x+θ) = \sin(x+ϕ)\cos(θ−ϕ)+\cos(x+ϕ)\sin(θ−ϕ) = \sin(x+ϕ+θ−ϕ)$$
(by the angle sum formula, and the fact that $\cot x=\frac{\cos x}{\sin x}$)
A: We have that $$\begin{align}\frac{\sin(x+\theta)}{\sin(x+\phi)}&=\frac{\sin\left((x\color{red}{+\phi})+(\theta\color{red}{-\phi})\right)}{\sin(x+\phi)} \\
&=\frac{\sin(x+\phi)\cos(\theta-\phi)+\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}& \\ \\
&=\frac{\color{blue}{\sin(x+\phi)}\cos(\theta-\phi)}{\color{blue}{\sin(x+\phi)}}+\frac{\sin(\theta(\phi)\color{green}{\cos(x+\phi)}}{\color{green}{\sin(x+\phi)}} \\ &=\cos(\theta-\phi)+\color{green}{\cot(x+\phi)}\sin(\theta-\phi)\end{align}$$
by the angle sum formula.
