1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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, 2019 at 8:58

1 Answer 1

Reset to default
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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, 2019 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.