Question

решите пожалуйста,срочно нада

Answer 1

$1) \ Sin\alpha =\dfrac{5}{13} \\\\\dfrac{\pi }{2} <\alpha <\pi \ \Rightarrow \ Cos\alpha <0\\\\Cos\alpha=-\sqrt{1-Sin^{2}\alpha } =-\sqrt{1-\Big(\dfrac{5}{13}\Big)^{2} } =-\sqrt{1-\dfrac{25}{169} } =-\sqrt{\dfrac{144}{169} } =-\dfrac{12}{13} \\\\tg\alpha =\dfrac{Sin\alpha }{Cos\alpha }=\dfrac{5}{13} :\Big(-\dfrac{12}{13} \Big)=-\dfrac{5}{13} \cdot\dfrac{13}{12} =\boxed{-\dfrac{5}{12} }$

$2) \ Sin\alpha =\dfrac{5}{13} \\\\\dfrac{\pi }{2} <\alpha <\pi \ \Rightarrow \ Cos\alpha <0\\\\Cos\alpha=-\sqrt{1-Sin^{2}\alpha } =-\sqrt{1-\Big(\dfrac{5}{13}\Big)^{2} } =-\sqrt{1-\dfrac{25}{169} } =-\sqrt{\dfrac{144}{169} } =-\dfrac{12}{13} \\\\Ctg\alpha =\dfrac{Cos\alpha }{Sin\alpha }=-\dfrac{12}{13} :\dfrac{5}{13} =-\dfrac{12}{13} \cdot\dfrac{13}{5} =\boxed{-\dfrac{12}{5} }$

$3) \ Ctgx=2\\\\\dfrac{(2+2Sinx)(1-Sinx)}{(1+Cosx)(2-2Cosx)} =\dfrac{2(1+Sinx)(1-Sinx)}{(1+Cosx)2(1-Cosx)} =\dfrac{1-Sin^{2} x}{1-Cos^{2} x} =\\\\\\=\dfrac{Cos^{2} x}{Sin^{2}x } =Ctg^{2} x=2^{2} =\boxed4$