Question

Помогите Решить, Дам 30 балов!

Answer 1

$2)\; \; \dfrac{1}{cosa}-sina\cdot tga=\dfrac{1}{cosa}-sina\cdot \dfrac{sina}{cosa}=\dfrac{1-sin^2a}{cosa}=\dfrac{cos^2a}{cosa}=cosa\\\\\\3)\; \; \dfrac{sin^2a-1}{cos^2a-1}+tga\cdot ctga=\dfrac{-cos^2a}{-sin^2a}+1=ctg^2a+1=\dfrac{1}{sin^2a}$

$4)\; \; cosa=-\dfrac{1}{5}\; \; ,\; \; a\in (180^\circ ;270^\circ )\; \; \Rightarrow \; \; tga>0\\\\1+tg^2a=\dfrac{1}{cos^2a}\; \; \to \; \; \; 1+tg^2a=\dfrac{1}{1/25}=25\; ,\; \; tg^2a=24\\\\tga=\sqrt{24}=2\sqrt6>0\\\\\\5a)\; \; cos105+cos75=2\cdot cos90\cdot cos15=0\\\\b)\; \; sin\dfrac{11}{12}\pi +sin\dfrac{5}{12}\pi =2\cdot sin\dfrac{4\pi }{3}\cdot cos\dfrac{\pi}{4}=2\cdot (-\dfrac{\sqrt3}{2})\cdot \dfrac{\sqrt2}{2}=-\dfrac{\sqrt6}{2}$

$c)\; \; cos\dfrac{11}{12}\pi -cos\dfrac{5}{12}\pi =-2\cdot sin\dfrac{4\pi }{3}\cdot sin\dfrac{\pi}{4}=2\cdot \dfrac{\sqrt3}{2}\cdot \dfrac{\sqrt2}{2}=\dfrac{\sqrt6}{2}\\\\\\d)\; \; cos15-sin15=cos15-cos(90-15)=cos15-cos75=\\\\=-2\cdot sin45\cdot sin(-60)=2\cdot \dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt2}{2}$

$e)\; \; tg22^\circ 30'+tg67^\circ 30'=\dfrac{sin90^\circ }{cos22^\circ 30'\cdot cos67^\circ 30'}=\dfrac{1}{0,5\cdot (cos90^\circ +cos45^\circ )}=\\\\\\=\dfrac{2}{0+\frac{\sqrt2}{2}}=\dfrac{4}{\sqrt2}=\dfrac{4\sqrt2}{2}=2\sqrt2$

$f)\; \; tg\frac{11}{12}\pi -tg\frac{5}{12}\pi =\dfrac{sin\frac{\pi }{2}}{cos\frac{11\pi }{12}\cdot cos\frac{5\pi}{12}}=\dfrac{1}{0,5\cdot (cos\frac{8\pi }{3}+cos\frac{\pi}{2})}=\dfrac{2}{-\frac{1}{2}+0}=-4$