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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?

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  • $\begingroup$ 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.) $\endgroup$ – Blue May 16 '19 at 8:58
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$$\frac{1+\csc^2A\sin^2C-\sin^2C}{1+\csc^2B\sin^2C-\sin^2C}=\frac{(\cos^2C+\csc^2A\sin^2C)\sec^2C}{(\cos^2C+\csc^2B\sin^2C)\sec^2C}=\frac{1+\csc^2A\tan^2C}{1+\csc^2B\tan^2C}$$

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  • $\begingroup$ 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. $\endgroup$ – Andrei Lenedin May 16 '19 at 6:51

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