Question

sinacosa/sin^2a-cos^2a esli
tga=3/2​

Answer 1

$\displaystyle \frac{sina*cosa}{sin^2-cos^2a}=\frac{\frac{2}{2}sina*cosa}{-(cos^2a-sin^2a)}= -\frac{1}{2}\frac{sin2a}{cos2a}=-\frac{1}{2}*tg2a=\\\\=-\frac{1}{2}*\frac{2tga}{1-tg^2a}\\\\ tga=\frac{3}{2}\\\\-\frac{1}{2}*\frac{2*\frac{3}{2}}{1-(\frac{3}{2})^2}=-\frac{1}{2}*\frac{3}{\frac{-5}{4}}=\frac{6}{5}$

Answer 2

Ответ:

1.2

Объяснение: