$\displaystyle \frac{1+tan^2[\frac{\pi}4-a]}{1-tan^2[\frac{\pi}4-a]}=cos 2A$ Prove this
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Assuming a = A? Have you tried to expand the LHS out with the compound angle formula for tan? Recall $\displaystyle \displaystyle \tan (a\pm b)= \frac{\tan a \pm \tan b}{1\mp \tan a \tan b}$
I tried this way but finally i get 1/sin2a.
Originally Posted by srirahulan $\displaystyle \frac{1+tan^2[\frac{\pi}4-a]}{1-tan^2[\frac{\pi}4-a]}=cos 2A$ Prove this As written that statement is false. To see that let $\displaystyle a=\frac{\pi}{4}$ If you change $\displaystyle \cos(2a)$ to $\displaystyle \csc(2a)$ then it is true.
yes,I can understand.
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