How do I prove $\frac{\sin x + \csc y}{\sin y + \csc x} = \csc y \sin x$?

Question

How do I prove $\frac{\sin x + \csc y}{\sin y + \csc x} = \csc y \sin x$?

I managed to get to $\frac{\sin^2x\sin y+1}{\sin^2y\sin x+1}$ from the left hand side and then got stuck.

Answer 1

$$\begin{align} \csc{(y)}\sin{(x)} &=\csc{(y)}\sin{(x)}\cdot\frac{\sin{(y)}+\csc{(x)}}{\sin{(y)}+\csc{(x)}}\\ &=\frac{\csc{(y)}\sin{(x)}\sin{(y)}+\csc{(y)}\sin{(x)}\csc{(x)}}{\sin{(y)}+\csc{(x)}}\\ &=\frac{\sin{(x)}+\csc{(y)}}{\sin{(y)}+\csc{(x)}}\\ \end{align}$$

Answer 2

If $\sin x=p,\sin y=q$

$\dfrac{p+\dfrac1q}{q+\dfrac1p}=\dfrac pq$ if $pq+1\ne0$

Answer 3

The left-hand side is given by $$\frac{\sin(x)(1+\sin(x))}{\sin(y)(1+\sin(x))}$$