Question

Срочноооо даю 40 балов!!!!!

Answer 1

$1)\; \; \frac{tg^2\phi -1}{tg^2\phi +1}+cos^2\phi =\frac{\frac{sin^2\phi }{cos^2\phi }-1}{\frac{sin^2\phi }{cos^2\phi }+1}+cos^2\phi =\frac{\frac{sin^2\phi -cos^2\phi }{cos^2\phi }}{\frac{sin^2\phi +cos^2\phi }{cos^2\phi }}+cos^2\phi =\\\\=\frac{sin^2\phi -cos^2\phi }{cos^2\phi }\cdot \frac{cos^2\phi }{1}+cos^2\phi =sin^2\phi -cos^2\phi +cos^2\phi =sin^2\phi$

$2)\; \; cosa=-0,8\\\\sin^2a=1-cos^2a\; \; \to \; \; \; sina=\pm \sqrt{1-cos^2a}\\\\\pi <a<\frac{3\pi }{2} \; \; \to \; \; sina<0\; \; \to \; \; sina=-\sqrt{1-cos^2a}=-\sqrt{1-0,64}=-0,6\\\\tga=\frac{sina}{cosa}=\frac{-0,6}{-0,8}=0,75\\\\ctga=\frac{1}{tga}=\frac{1}{0,75}=\frac{1}{3/4}=\frac{4}{3}$

$3)\; \; ctga\cdot sin^2a=\frac{cosa}{sina}\cdot sin^2a=cosa\cdot sina=\frac{1}{2}\, sin2a\; ;\\\\(tga+ctga)^{-1}=\Big (\frac{sina}{cosa}+\frac{cosa}{sina}\Big )^{-1}=\Big (\frac{sin^2a+cos^2a}{sina\cdot cosa}\Big )^{-1}=\\\\=\Big (\frac{1}{\frac{1}{2}\cdot sin2a}\Big )^{-1}=\Big (\frac{2}{sin2a}\Big )^{-1}=\frac{sin2a\; ;\\\\\frac{x}{y} }{2}=\frac{1}{2}\, sin2a\; ; \\\\\\\frac{1}{2}\, sin2a=\frac{1}{2}\, sin2a\; \; \Rightarrow \; \; \; ctga\cdot sin^2a=(tga+ctga)^{-1}$

Answer 2

рЕШЕНИЯ И ОТВЕТ ВО ВЛОЖЕНИИ