Question

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Answer 1

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

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

$64)\ \ tga+ctga=\dfrac{2}{sin2a} =\sqrt{125}=5\sqrt5\ \ ,\ \ \dfrac{\pi}{4}<a<\dfrac{\pi }{2}\\\\sin2a=\dfrac{2}{tga+ctga}=\dfrac{2}{5\sqrt5}\ \ ,\\\\\\\dfrac{\pi}{2}<2a<\pi \ \ \to \ \ \ cos2a<0\\\\cos2a=-\sqrt{1-sin^22a}=-\sqrt{1-\dfrac{4}{125}}=-\dfrac{11}{5\sqrt5}\ \ \to \\\\\\tg2a=\dfrac{sin2a}{cos2a}=-\dfrac{2}{11}<0\ \ \ (\ \dfrac{\pi}{2}<2a<\pi \ )\\\\\\ctga-tga=\dfrac{2}{tg2a}=\dfrac{2}{\frac{2}{-11}}=-11$