Question

Подробно решите пожалуйста ​

Answer 1

Ответ:

$\boxed{\ \displaystyle ctg(x-y)=\dfrac{1}{tg(x-y)}=\frac{1+tgx\cdot tgy}{tgx-tgy}\ }\\\\\\x=2\alpha \ ,\ y=2\beta \ \ \Rightarrow \ \ \ ctg(2\alpha -2\beta )=\dfrac{1+tg2\alpha \cdot tg2\beta }{tg2\alpha -tg2\beta }\\\\\\\\tg\alpha =\dfrac{1}{4}\ \ ,\ \ tg\beta =2\\\\tg2\alpha =\dfrac{2tg\alpha }{1-tg^2\alpha }=\dfrac{2\cdot \dfrac{1}{4}}{1-\dfrac{1}{16}}=\dfrac{\dfrac{1}{2}}{\dfrac{15}{16}}=\dfrac{16}{2\cdot 15}=\dfrac{8}{15}$

$tg2\beta =\dfrac{2tg\beta }{1-tg^2\beta }=\dfrac{2\cdot 2}{1-4}=-\dfrac{4}{3}\\\\\\ctg(2\alpha -2\beta )=\dfrac{1+\dfrac{8}{15}\cdot (-\dfrac{4}{3})}{\dfrac{8}{15}+\dfrac{4}{3}}=\dfrac{\dfrac{13}{45}}{\dfrac{28}{15}}=\dfrac{13\cdot 15}{45\cdot 28}=\dfrac{13}{84}$