Question

ДАЮ 15 БАЛОВ!!!!
Попрошу с об'яснением и без обмана​

Answer 1

$63)\ \ tga-ctga=\sqrt{12}\ \ ,\ \ \ \dfrac{3\pi}{4}<a<\pi\\\\\\\\\star\ tg2a=\dfrac{sin2a}{cos2a}=\dfrac{2sina\cdot cosa}{cos^2a-sin^2a}=\dfrac{2}{\frac{cos^2a}{sina\cdot cosa}-\frac{sin^2a}{sina\cdot cosa}}=\\\\\\=\dfrac{2}{\frac{cosa}{sina}-\frac{sina}{cosa}}=\dfrac{2}{ctga-tga}\ \ \ \Rightarrow \ \ \ ctga-tga=\dfrac{2}{tg2a\ \ }\star \\\\\\tga-ctga=-\dfrac{2}{tg2a}=\sqrt{12}=2\sqrt3\ \ \to \ \ \ tg2a=-\dfrac{1}{\sqrt3}<0\ \ ,\ (\dfrac{3\pi}{2}<2a< 2\pi )$

$tga+ctga=\dfrac{sina}{cosa}+\dfrac{cosa}{sina}=\dfrac{sin^2a+cos^2a}{sina\cdot cosa}=\dfrac{1}{\frac{1}{2}\, sin2a}=\dfrac{2}{sin2a}\\\\\\\star \ \ 1+ctg^22a=\dfrac{1}{sin^22a}\ \ \to \ \ 1+\dfrac{1}{tg^22a}=\dfrac{1}{sin^22a}\ \ ,\ \ \dfrac{1+tg^22a}{tg^22a}=\dfrac{1}{sin^22a}\ \ ,\\\\\\sin^22a=\dfrac{tg^22a}{1+tg^22a}=\dfrac{\frac{1}{3}}{1+\frac{1}{3}}=\dfrac{1}{4}\ \ \to \ \ \ sin2a=-\dfrac{1}{2}<0\ \ (\, \dfrac{3\pi}{2}<2a<2\pi \, )\\\\\\tga+ctga=\dfrac{2}{-\frac{1}{2}}=-4$

$62)\ \ tga-ctga=-\sqrt5\ \ ,\ \ \dfrac{\pi}{2}<a<\dfrac{3\pi }{4}\\\\tg2a=\dfrac{2}{ctga-tga}=\dfrac{2}{\sqrt5}>0\ \ ,\ \ \ \pi <2a<\dfrac{3\pi}{2}\\\\\\sin^22a=\dfrac{tg^22a}{1+tg^22a}=\dfrac{\frac{4}{5}}{1+\frac{4}{5}}=\dfrac{4}{9}\ \ \to \\\\\\sin2a=-\dfrac{2}{3}<0\ \ \ (\ \pi <2a<\dfrac{3\pi }{2}\ )\\\\\\tga+ctga=\dfrac{2}{sin2a}=\dfrac{2}{-\frac{2}{3}}=-3$