Question

даю 30 балів срочно!!! ​

Answer 1

$\displaystyle\bf\\\frac{1-Cos^{2}\alpha }{Sin\alpha Cos\alpha } +Sin^{2}\alpha (tg\alpha +Ctg\alpha )=\\\\\\= \frac{Sin^{2}\alpha }{Sin\alpha Cos\alpha } +Sin^{2}\alpha\Big (\frac{Sin\alpha }{Cos\alpha } +\frac{Cos\alpha }{Sin\alpha } \Big )=\\\\\\=\frac{Sin\alpha }{ Cos\alpha } +Sin^{2}\alpha\Big (\frac{Sin^{2} \alpha +Cos^{2} \alpha }{Sin\alpha \cdot Cos\alpha } \Big )= \\\\\\=\frac{Sin\alpha }{ Cos\alpha } +Sin^{2}\alpha\cdot\frac{1 }{Sin\alpha \cdot Cos\alpha } =$

$\displaystyle\bf\\=\frac{Sin\alpha }{ Cos\alpha } +\frac{Sin\alpha }{Cos\alpha } =2\cdot\frac{Sin\alpha }{Cos\alpha } =2tg\alpha \\\\\\2tg\alpha =2tg\alpha$

Тождество доказано

Answer 2

Відповідь:

Пояснення:

$\frac{1-cos^2\alpha }{sin\alpha cos\alpha } +sin^2\alpha(tg\alpha +ctg\alpha )=2tg\alpha ;$

$\frac{sin^2\alpha }{sin\alpha cos\alpha } +sin^2\alpha (\frac{sin\alpha }{cos\alpha } +\frac{cos\alpha }{sin\alpha } )=2tg\alpha ;$

$\frac{sin\alpha }{ cos\alpha } +sin^2\alpha (\frac{sin^2\alpha +cos^2\alpha }{sin\alpha cos\alpha } } )=2tg\alpha ;$

$tg\alpha +sin^2\alpha \frac{1 }{sin\alpha cos\alpha } } =2tg\alpha ;$

$tg\alpha +\frac{sin\alpha }{cos\alpha } =2tg\alpha$

$tg\alpha +tg\alpha =2tg\alpha ;$

$2tg\alpha =2tg\alpha$