Question

1) ДОКАЖИТЕ ТОЖДЕСТВО: cos(a-90)+sin(a-180)+tg^2(180-a)+ctg^2(a-180)=tg^2(a)+ctg^2(a)
2)sin(t)=5/13 pi>t>pi/2, найти cos(t),tg^2(t),sin^2(t)

Answer 1

cos(α-90)=cos(-(90-α))=cos(90-α)=sinα
sin(α-180)=sin(-(180-α))=-sin(180-α)=-sinα
tg²(180-α)=(tg(180-α))²=(-tgα)²=tg²α
ctg²(α-180)=(ctg(-(180-α)))²=(-ctg(180-α))²=(ctgα)²=ctg²α

cos(α-90)+sin(α-180)+tg²(180-α)+ctg²(α-180)=sinα-sinα+tg²α+ctg²α=tg²α+ctg²α
tg²α+ctgα=tg²α+ctg²α

2.  sin²t=(5/13)²,  sin²t=25/169 
sin²t+cos²t=1 
cos²t=1-(5/13)², cos²t=144/169
cost=+-12/13, π/2<t<π, ⇒cost<0
cost=-12/13
tg²t=sin²t/cos²t, tg²t=(25/169)/(144/169).
tg²t=25/144

Answer 2

$1)cos(a-90)+sin(a-180)+tg ^{2} (180-a)+ctg ^{2} (a-180)= \\ =tg ^{2} a+ctg ^{2} a \\ \\ cos(a-90)+sin(a-180)+tg ^{2} (180-a)+ctg ^{2} (a-180)= \\ =cos(90-a)-sin(180-a)+tg^{2} a-ctg ^{2} (180-a)= \\ =sina-sina+tg ^{2} a+ctg ^{2} a=tg ^{2} a+ctg ^{2} a$



$2)sint= \frac{5}{13} ;t\in ( \frac{ \pi }{2} ; \pi ) \tot\in II \\ \\ cost=- \sqrt{1-sin ^{2} t} =- \sqrt{1- (\frac{5}{13} ) ^{2} } =- \sqrt{1- \frac{25}{169} } =- \sqrt{ \frac{169-25}{169} } = \\ \\ =- \sqrt{ \frac{144}{169} } =- \frac{12}{13} \\ \\ sin ^{2}t = \frac{25}{169} \\ \\ cos ^{2} t= \frac{144}{169} \\ \\ tg ^{2} t= \frac{sin ^{2} t}{cos^{2} t} = \frac{25}{169} : \frac{144}{169} = \frac{25*169}{169*144} = \frac{25}{144}$