Одно любое задание (но желательно все)

7.
$cosx= -\sqrt{1- \frac{a^2}{a^2+b^2} } =- \frac{b}{ \sqrt{a^2+b^2} } \\ tgx= \frac{sinx}{cosx} =- \frac{a}{b} \\ ctgx= \frac{1}{tgx}=- \frac{b}{a}$

$\frac{5cosa+6sina}{3sina-7cosa} = \frac{ \frac{5cosa+6sina}{cosa} }{ \frac{3sina-7cosa}{cosa} } = \frac{5+6tga}{3tga-7} = \frac{5+0,5*6}{3*0,5-7} = \frac{8}{-5,5} =- \frac{80}{55} =\\- \frac{16}{11} =-1 \frac{5}{11}$

$\frac{3sin^2a+12sinacosa+cos^2a}{sin^2a+sinacosa-2cos^2a} = \frac{ \frac{3sin^2a+12sinacosa+cos^2a}{cos^2a} }{ \frac{sin^2a+sinacosa-2cos^2a}{cos^2a} } = \frac{3tg^2a+12tga+1}{tg^2a+tga-2} =\\ = \frac{3*4+12*2+1}{4+2-2} = \frac{37}{4} =9,25$

$\sqrt{ \frac{1+sina}{1-sina} } - \sqrt{ \frac{1-sina}{1+sina}} = \sqrt{ \frac{(1+sina)^2}{(1-sina)(1+sina)} } - \sqrt{\frac{(1-sina)^2}{(1-sina)(1+sina)} } =\\ = \frac{1+sina}{ \sqrt{1-sin^2a} } - \frac{1-sina}{ \sqrt{1-sin^2a} } = \frac{2sina}{ \sqrt{cos^2a} } = \frac{2sina}{-cosa} =-2tga$