Question

ТРИГОНОМЕТРИЯ. ПОМОГИТЕ, ПОЖАЛУЙСТА. ​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{Cos10^\circ-1}{Sin10^\circ } =\frac{1-2Sin^{2} 5^\circ-1 }{2Sin5^\circ Cos5^\circ } =\frac{-2Sin^{2} 5^\circ}{2Sin5^\circ Cos5^\circ } =-\frac{Sin5^\circ }{Cos5^\circ} =-tg5^\circ\\\\\\2)\\\\\frac{Ctg(7\pi -\beta )}{Cos\Big(\frac{11\pi }{2} +\beta \Big)} =\frac{Ctg(-\beta )}{Cos\Big[6\pi -(\frac{\pi }{2} -\beta )\Big]} =\frac{-Ctg\beta }{Cos\Big(\frac{\pi }{2}-\beta\Big) } =\\\\\\=-\frac{Cos\beta }{Sin\beta } :Sin\beta =-\frac{Cos\beta }{Sin^{2} \beta }$