Question

Здравствуйте! Пожалуйста помогите.

Answer 1

1)
a)
$\frac{1-sin \alpha }{cos \alpha } +tg \alpha = \frac{1-sin \alpha }{cos \alpha } + \frac{sin \alpha }{cos \alpha } = \frac{1-sin \alpha +sin \alpha }{cos \alpha } = \frac{1}{cos \alpha }$
б)
$ctg \beta - \frac{cos \beta -1}{sin \beta } = \frac{cos \beta }{sin \beta } - \frac{cos \beta -1}{sin \beta }=\frac{cos \beta -(cos \beta -1)}{sin \beta }=\frac{cos \beta -cos \beta +1}{sin \beta }= \frac{1}{sin \beta }$

2)
a)
$\frac{1+2sin \beta cos \beta }{(cos \beta +sin \beta )^2}= \frac{1+2sin \beta cos \beta }{cos^2 \beta +sin^2\beta +2sin \beta cos \beta }= \frac{1+2sin \beta cos \beta }{1+2sin \beta cos \beta }=1$
б)
$\frac{1}{1+tg^2 \beta }+ \frac{1}{1+ctg^2 \beta } = \frac{1}{ \frac{1}{cos^2 \beta } } + \frac{1}{ \frac{1}{sin^2 \beta } }=cos^2 \beta +sin^2 \beta =1$

$cos^2x+sin^2x=1$
$tgx= \frac{sinx}{cosx}$
$ctgx= \frac{cosx}{sinx}$
$1+tg^2x= \frac{1}{cos^2x}$
$1+ctg^2x= \frac{1}{sin^2x}$