Question

срочно
даю 50 баллов​

Answer 1

Ответ:

Формулы      $\boxed{\ cos2x=cos^2x-sin^2x\ ,\ \ ctgx=\dfrac{cosx}{sinx}\ }$   .

$\displaystyle ctg^2a-1=\frac{cos2a}{sin^2a}\\\\\\ctg^2a-1=\frac{cos^2a}{sin^2a}-1=\frac{cos^2a-sin^2a}{sin^2a}=\frac{cos2a}{sin^2a}\\\\\\\frac{cos2a}{sin^2a}=\frac{cos2a}{sin^2a}$