Решить 422(1,3,5) 423 (1,3,5,7) большое спасибо!

422 [latex]1) ctg \beta - \frac{cos \beta }{sin \beta } + \frac{1}{sin \beta } =ctg \beta -ctg \beta + \frac{1}{sin \beta } = \frac{1}{sin \beta }[/latex] [latex]3) \frac{1-ctgy}{tgy-1}= \frac{1- \frac{cosy}{siny} }{ \frac{siny}{cosy} -1}= \frac{ \frac{siny-cosy}{siny} }{ \frac{siny-cosy}{cosy}}= \frac{cosy}{siny} =ctgy[/latex] [latex]5)tg^2 \alpha (sin^2 \alpha -cos^ \alpha -sin^ \alpha )=- \frac{sin^2 \alpha }{cos^2 \alpha } *cos^2 \alpha =-sin^2 \alpha [/latex] 423 [latex]1) -tg \alpha *cos \alpha +sin \alpha =-sin \alpha +sin \alpha =0[/latex] [latex]3) - \frac{ctg \beta *sin \beta }{cos \beta } =- \frac{cos \beta }{cos \beta }=-1[/latex] [latex]5) ctg \alpha (-sin \alpha )-cos \alpha =-cos \alpha -cos \alpha =-2cos \alpha [/latex] [latex]7) \frac{1-sin^2(-y)}{cosy}= \frac{cis^2y}{cosy}=cosy [/latex]