Prove $$\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}=\frac{1+\cot^2A\sin^2C}{1+\cot^2B\sin^2C}$$
I chose to manipulate the left hand side of the equation, by firstly replacing $\cot^2A$ with $\csc^2A-1$ according to the identities. After doing the same with the denominator, I'm left with,
$$\text{RHS}=\frac{1+(\csc^2A-1)\sin^2C}{1+(\csc^2B-1)\sin^2C} =\frac{1+\csc^2A\sin^2C-\sin^2C}{1+\csc^2B\sin^2C-\sin^2C}$$
And, by contracting $1-\sin^2C$ to $\cos^2C$, on both top and bottom, I can't think of applying anything else.
Anyone know how to continue?