Question

помогите пожалуйста (

Answer 1

Объяснение:

фото.........

...

.......

Answer 2

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

$\displaystyle\bf\\3)\\\\\frac{Sin^{2} \alpha }{Cos^{2} \alpha -1} =-\frac{Sin^{2} \alpha }{1-Cos^{2} \alpha }=-\frac{Sin^{2} \alpha }{Sin^{2} \alpha }=-1\\\\\\4)\\\\1+Cos^{2} \alpha -Sin^{2}\alpha =\underbrace{(1-Sin^{2} \alpha ) }_{Cos^{2}\alpha } +Cos^{2} \alpha =Cos^{2} \alpha +Cos^{2} \alpha =2Cos^{2} \alpha$