Question

Вычислите tgB если ...
Решите с подробным объяснением

Answer 1

Answer 2

$tg\beta=1$

Объяснение:

$\frac{cos(a+\beta)}{cos(a-\beta)}=\frac{2}{3},\; \; \; ctga=5\; \; \; tg\beta=?\\\\\\\frac{cos(a+\beta)}{cos(a-\beta)}=\frac{cosacos\beta-sinasin\beta }{cosacos\beta+sinasin\beta}=\frac{\frac{cosacos\beta}{cosacos\beta}-\frac{sinasin\beta}{cosacos\beta}}{\frac{cosacos\beta}{cosacos\beta}+\frac{sinasin\beta}{cosacos\beta}}=\frac{1-tgatg\beta}{1+tgatg\beta}\\\\\frac{cos(a+\beta)}{cos(a-\beta)}=\frac{2}{3}\; \; =\; \;\frac{1-tgatg\beta}{1+tgatg\beta}=\frac{2}{3}$

$tga=\frac{1}{ctga}=\frac{1}{5}\\\\\frac{1-\frac{1}{5}tg\beta}{1+\frac{1}{5}tg\beta}=\frac{2}{3}\\\\3-\frac{3}{5}tg\beta=2+\frac{2}{5}tg\beta\\\\\ 3-2=\frac{2}{5}tg\beta+\frac{3}{5}tg\beta\\\\1=\frac{5}{5}tg\beta\\\\1=tg\beta\\\\tg\beta=1$