Question

ctg(b-a) (tga-tgb) +tgatgb+cos2y​

Answer 1

Ответ:

$\cos 2y-1$

Объяснение:

$\cot(b-a)(\tan a-\tan b)+\tan a*\tan b+\cos 2y=\\=\frac{\cos(b-a)}{\sin(b-a)}(\frac{\sin a}{\cos a}-\frac{\sin b}{\cos b})+\frac{\sin a\sin b}{\cos a\cos b}+\cos 2y=\\=\frac{\cos a\cos b+ \sin a\sin b}{\sin b \cos a-\cos b\sin a}\frac{\sin a\cos b-\cos a\sin b}{\cos a\cos b}+\frac{\sin a\sin b}{\cos a\cos b}+\cos 2y=\\=-\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b}+\frac{\sin a\sin b}{\cos a\cos b}+\cos 2y=\\=-\frac{\cos a\cos b}{\cos a\cos b}+\cos 2y=\cos 2y-1$