Trigonometry trig fix

[Sudharaka]

srirahulan's question titled "trig fix" from Math Help Forum,

Prove that, \[\frac{1+\tan^{2}(\frac{\pi}{4}-A)}{1-\tan^{2}(\frac{\pi}{4}-A)}=\cos 2A\]

Hi srirahulan,

Consider the left hand side of the equation.

\begin{eqnarray}

\frac{1+\tan^{2}(\frac{\pi}{4}-A)}{1-\tan^{2}(\frac{\pi}{4}-A)}&=&\frac{\cos^{2}(\frac{\pi}{4}-A)+\sin^{2}(\frac{\pi}{4}-A)}{\cos^{2}(\frac{\pi}{4}-A)-\sin^{2}(\frac{\pi}{4}-A)}\\

&=&\frac{1}{\cos 2(\frac{\pi}{4}-A)}\\

&=&\frac{1}{\cos (\frac{\pi}{2}-2A)}\\

&=&\frac{1}{\sin 2A}\\

\end{eqnarray}

\[\therefore \frac{1+\tan^{2}(\frac{\pi}{4}-A)}{1-\tan^{2}(\frac{\pi}{4}-A)} = \csc 2A\]

So I think there is either a mistake in the question or a typo on your part.