$\displaystyle \frac{1+tan^2[\frac{\pi}4-a]}{1-tan^2[\frac{\pi}4-a]}=cos 2A$Prove this

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- May 27th 2012, 08:35 PMsrirahulantrig fix
$\displaystyle \frac{1+tan^2[\frac{\pi}4-a]}{1-tan^2[\frac{\pi}4-a]}=cos 2A$

**Prove this** - May 27th 2012, 09:08 PMpickslidesRe: trig fix
Assuming a = A?

Have you tried to expand the LHS out with the compound angle formula for tan?

Recall $\displaystyle \displaystyle \tan (a\pm b)= \frac{\tan a \pm \tan b}{1\mp \tan a \tan b}$ - May 28th 2012, 06:40 AMsrirahulanRe: trig fix
I tried this way but finally i get 1/sin2a.

- May 28th 2012, 07:25 AMPlatoRe: trig fix
- May 29th 2012, 05:01 PMsrirahulanRe: trig fix
yes,I can understand.