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- #1
- Feb 5, 2012
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srirahulan's question titled "trig fix" from Math Help Forum,
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.
Hi srirahulan,Prove that, \[\frac{1+\tan^{2}(\frac{\pi}{4}-A)}{1-\tan^{2}(\frac{\pi}{4}-A)}=\cos 2A\]
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.