# trig fix

• May 27th 2012, 08:35 PM
srirahulan
trig fix
$\frac{1+tan^2[\frac{\pi}4-a]}{1-tan^2[\frac{\pi}4-a]}=cos 2A$ Prove this
• May 27th 2012, 09:08 PM
pickslides
Re: trig fix
Assuming a = A?

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

Recall $\displaystyle \tan (a\pm b)= \frac{\tan a \pm \tan b}{1\mp \tan a \tan b}$
• May 28th 2012, 06:40 AM
srirahulan
Re: trig fix
I tried this way but finally i get 1/sin2a.
• May 28th 2012, 07:25 AM
Plato
Re: trig fix
Quote:

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

As written that statement is false.
To see that let $a=\frac{\pi}{4}$

If you change $\cos(2a)$ to $\csc(2a)$ then it is true.
• May 29th 2012, 05:01 PM
srirahulan
Re: trig fix
yes,I can understand.