Question

Задание 4 Б) Пожалуйста! Вычислите!

Answer 1

$1+tg^2a = \frac{1}{cos^2a} \\ \\ 1+\frac{4}{9} = \frac{1}{x} \\ \\ \frac{13}{9} =\frac{1}{a} \\ \\ a=\frac{9}{13} \\ \\ cos a = \frac{3}{\sqrt{13} } \\ \\ \\ sin^2a = 1-cos^2a \\ \\ sin^2a = 1-9/13 = 4/13\\ \\ sina = 2/\sqrt{13} \\ \\ \\ ctga = cosa/sina \\ \\ ctga = \frac{3*\sqrt{13} }{\sqrt{13} *2} = 3/2 = 1.5$