Question

помогите пожалуйста)
465(1) и 461(2)

Answer 1

$\frac{1+\cos42}{1-\cos42} = \cfrac{ \frac{1+\cos42}{2} }{ \frac{1-\cos42}{2} } = \frac{ \cos^221 }{\sin^221 } =\mathrm{ctg}^221$

$\cos \alpha =- \sqrt{1-\sin^2 \alpha } =- \sqrt{1-(-\frac{12}{13} )^2} =- \frac{5}{13} \\\\ \cos2 \alpha =\cos^2 \alpha -\sin^2 \alpha =(- \frac{5}{13} )^2-(- \frac{12}{13} )^2= \frac{25}{169} -\frac{144}{169}=- \frac{119}{169} \\\\ \sin2 \alpha =2\sin \alpha \cos \alpha =2\cdot(- \frac{5}{13} )\cdot(- \frac{12}{13} )= \frac{120}{169}$