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

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?

• I formatted the equations from your images. I took the liberty of converting $\tan^2A$ into $\tan^2C$. (The identity isn't true otherwise; plus, the $A$ seems to have turned into a $C$ in the work shown, which makes the original $A$ look even more like a typographical error.) – Blue May 16 '19 at 8:58

$$\frac{1+\csc^2A\sin^2C-\sin^2C}{1+\csc^2B\sin^2C-\sin^2C}=\frac{(\cos^2C+\csc^2A\sin^2C)\sec^2C}{(\cos^2C+\csc^2B\sin^2C)\sec^2C}=\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}$$
• I believe my op had an error, since $tan^2A$ in the question was meant to be $tan^2C$. That is the only way I can see your answer as correct. – Andrei Lenedin May 16 '19 at 6:51