Question

Помогите решить Ббббббббб​

Answer 1

Ответ:

$\boxed{\ sin^2x+cos^2x=1\ \Rightarrow \ cos^2x=1-sin^2x\ }\\\\\boxed{\ \ tgx=\dfrac{sinx}{cosx}\ ;\ ctgx=\dfrac{cosx}{sinx}\ ;\ tgx\cdot ctgx=1\ }$

$\displaystyle a)\ \frac{cos\beta }{1-sin\beta }-\frac{cos\beta }{1+sin\beta }=\frac{cos\beta (1+sin\beta )-cos\beta (1-sin\beta )}{(1-sin\beta )(1+sin\beta )}=\frac{2\, sin\beta \cdot cos\beta }{1-sin^2\beta }=\\\\\\=\frac{2\, sin\beta \cdot cos\beta }{cos^2\beta }=\frac{2\, sin\beta }{cos\beta }=2\, tg\beta$

$\displaystyle b)\ \ sin^2x-\underbrace{tg2\alpha \cdot ctg2\alpha }_{1}=sin^2x-1=-(1-sin^2x)=-cos^2x$