Докажите тождество
1-2sin^2A/(1+sin2A) =1-tgA/1+tgA

\displaystyle \frac{1-2\sin^2\alpha}{1+\sin2\alpha} = \frac{\cos 2\alpha}{1+\sin2\alpha} = \frac{(1-\tan^2\alpha)/(1+\tan^2\alpha)}{1+2\tan\alpha/(1+\tan^2\alpha)} = \\\\ = \frac{1-\tan^2\alpha}{(1+\tan\alpha)^2} = \frac{1-\tan\alpha}{1+\tan\alpha}