How to prove in this trig problem i have to prove this
$$\frac{\cos 3x}{\sin 2x \sin 4x}+\frac{\cos 5x}{\sin 4x \sin 6x}+\frac{\cos 7x}{\sin 6x \sin 8x}+\frac{\cos 9x}{\sin 8x \sin 10x} = \frac{1}{2}\csc x(\csc 2x - \csc 10x)$$
i tried taking lcm but does not leads to anything. i believe i have to write numerator as something in terms of denominator which i do not know how. Help. 
Thanks

The right side looks like a result of a telescopic sum, so we'll try to prove
$$ \frac{\cos(nx)}{\sin((n-1)x)\sin((n+1)x)}=\frac{1}{2}\csc(x)(\csc((n-1)x)-\csc((n+1)x)) $$
Why this? It may not be true, but if it is, we have
$$\begin{align*}
& \frac{\cos 3x}{\sin 2x\sin 4x} + \frac{\cos 5x}{\sin 4x\sin 6x} +\frac{\cos 7x}{\sin 6x\sin8x} +\frac{\cos 9x}{\sin 8x\sin10x} \\
=& \frac{1}{2}\csc x\left(\csc 2x - \csc 4x + \csc 4x - \csc 6x + \csc 6x - \csc 8x + \csc 8x - \csc 10x \right) \\
=& \frac{1}{2}\csc x (\csc 2x-\csc 10x) \\
\end{align*}$$
So, we'll start operating the right side. We have
$$\begin{align*}
& \frac{1}{2}\csc(x)(\csc((n-1)x)-\csc((n+1)x)) \\
=& \frac{1}{2\sin (x)}\left( \frac{1}{\sin((n-1)x)}-\frac{1}{\sin((n+1)x)} \right) \\
=& \frac{1}{2\sin(x)}\cdot  \frac{\sin((n+1)x)-\sin((n-1)x)}{\sin((n-1)x)\sin((n+1)x)}  \\
=& \frac{\sin((n+1)x)-\sin((n-1)x)}{2\sin(x)\sin((n-1)x)\sin((n+1)x)}  \\
=& \frac{(\sin(nx)\cos(x)+\sin(x)\cos(nx))-(\sin(nx)\cos(x)-\sin(x)\cos(nx))}{2\sin(x)\sin((n-1)x)\sin((n+1)x)} \\
=& \frac{2\sin(x)\cos(nx)}{2\sin(x)\sin((n-1)x)\sin((n+1)x)} \\
=& \frac{\cos(nx)}{\sin((n-1)x)\sin((n+1)x)} 
\end{align*}$$
And we're ready
