Question

Здравствуйте, помогите пожалуйста ​

Answer 1

Ответ:

$1)\ \ cosa=\dfrac{12}{13}\\\\\\0<a<\dfrac{\pi }{2}\ \ \to \ \ sina>0\ ,\ cosa>0\ ,\ tga>0\ ,\ ctga>0\\\\\\sina=+\sqrt{1-cos^2a}=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}\\\\\\tga=\dfrac{sina}{cosa}=\dfrac{5/13}{12/13}=\dfrac{5}{12}\\\\\\ctga=\dfrac{1}{tga}=\dfrac{12}{5}=2,4$

$2)\ \ sina=-\dfrac{\sqrt3}{4}\\\\\\\pi <a<\dfrac{3\pi }{2}\ \ \to \ \ sina<0\ ,\ cosa<0\ ,\ tga>0\ ,\ ctga>0\\\\\\cosa=-\sqrt{1-sin^2a}=-\sqrt{1-\dfrac{3}{16}}=-\sqrt{\dfrac{13}{16}}=-\dfrac{\sqrt{13}}{4}\\\\\\tga=\dfrac{sina}{cosa}=\dfrac{\sqrt3}{\sqrt{13}}=\sqrt{\dfrac{3}{13}}\\\\\\ctga=\dfrac{1}{tga}=\sqrt{\dfrac{13}{3}}$

$3)\ \ tga=-\dfrac{1}{3}\\\\\\\dfrac{3\pi}{2}<a<2\pi \ \ \ \to \ \ \ sina<0\ ,\ cosa>0\ ,\ tga<0\ ,\ ctga<0\\\\\\1+tg^2a=\dfrac{1}{cos^2a}\ \ ,\ \ cos^2a=\dfrac{1}{1+tg^2a}\\\\\\cosa=+\sqrt{\dfrac{1}{1+\frac{1}{9}}}=\dfrac{3}{\sqrt{10}}\\\\\\sina=-\sqrt{1-cos^2a}=-\sqrt{1-\dfrac{9}{10}}=-\sqrt{\dfrac{1}{10}}=-\dfrac{1}{\sqrt{10}}\\\\\\ctga=\dfrac{1}{tga}=-\dfrac{3}{1}=-3$

$4)\ \ ctga=-7\\\\\\\dfrac{\pi}{2}<a<\pi \ \ \ \to \ \ \ sina>0\ ,\ cosa<0\ ,\ tga<0\ ,\ ctga<0\\\\\\1+ctg^2a=\dfrac{1}{sin^2a}\ \ ,\ \ sin^2a=\dfrac{1}{1+ctg^2a}\\\\\\sina=+\sqrt{\dfrac{1}{1+49}}=\dfrac{1}{\sqrt{50}}=\dfrac{1}{5\sqrt2}\\\\\\cosa=-\sqrt{1-sin^2a}=-\sqrt{1-\dfrac{1}{50}}=-\sqrt{\dfrac{49}{50}}=-\dfrac{7}{\sqrt{50}}=-\dfrac{7}{5\sqrt2}\\\\\\tga=\dfrac{1}{ctga}=-\dfrac{1}{7}$