Proving $\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}=\frac{1+\cot^2A\sin^2C}{1+\cot^2B\sin^2C}$

Question

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?

Answer 1

$$\frac{1+\csc^2A\sin^2C-\sin^2C}{1+\csc^2B\sin^2C-\sin^2C}=\frac{(\cos^2C+\csc^2A\sin^2C)\sec^2C}{(\cos^2C+\csc^2B\sin^2C)\sec^2C}=\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}$$

Answer 2

Convert $tan^2(C)$ to $\frac{sin^2(C)}{cos^2(C)}$ $$\frac{1+\frac{\csc^2A\sin^2C}{cos^2C}}{1+\frac{\csc^2B\sin^2C}{cos^2C}}$$ Multiply all by $cos^2(C)$ $$\frac{cos^2C+\csc^2A\sin^2C}{cos^2C+\csc^2B\sin^2C}$$ Convert $cos^2C$ to $1-sin^2C$ $$\frac{1-sin^2C+\csc^2A\sin^2C}{1-sin^2C+\csc^2B\sin^2C}$$ Rearrange $$\frac{1+\csc^2A\sin^2C-sin^2C}{1+\csc^2B\sin^2C-sin^2C}$$ Factor out $sin^2C$ $$\frac{1+sin^2C(csc^2A-1)}{1+sin^2C(csc^2B-1)}$$ Use $csc^2x=cot^2x+1$ $$\frac{1+sin^2C(cot^2A+1-1)}{1+sin^2C(cot^2B+1-1)}$$ Simplify to RHS $$\frac{1+sin^2Ccot^2A}{1+sin^2Ccot^2B}$$