Question

вычислить : (кв корень(1-Cos a) / ка корень(1+ cos a )) + ( кв корень ( 1+Cos a) / квадратный корень ( 1 - cos a )) если sin a = -2/5

Answer 1

$sina=-\frac{2}{5}\\\\\frac{ \sqrt{1-cosa} }{ \sqrt{1+cosa}}+\frac{\sqrt{1+cosa}}{\sqrt{1-cosa}}=\frac{\sqrt{2sin^2 \frac{a}{2}}}{\sqrt{2cos^2\frac{a}{2}}}+\frac{\sqrt{2cos^2\frac{a}{2}}}{\sqrt{2sin^2\frac{a}{2}}}=\sqrt{tg^2\frac{a}{2}}+\sqrt{ctg^2\frac{a}{2}}=\\\\=|tg\frac{a}{2}|+|ctg\frac{a}{2}|=| \frac{sina}{1+cosa}|+| \frac{sina}{1-cosa}|=Q\\\\\star \; \; cos^2a=1-sin^2a=1- \frac{4}{25} = \frac{24}{25}\; ,\; \;cos \frac{a}{2}=\pm \frac{\sqrt{24}}{5}\; \; \star$

$1)\; \; sina=-\frac{2}{5}\; ,\; cosa= \frac{\sqrt{24}}{5}\\\\Q=\Big |\frac{-2/5}{1+\frac{\sqrt{24}}{5}}\Big |+\Big |\frac{-2/5}{1-\frac{\sqrt{24}}{5}}\Big |=\frac{2\cdot 5}{5\cdot (5+\sqrt{24})}+\frac{2\cdot 5}{5\cdot (5-\sqrt{24})}=\\\\=\frac{2\cdot (5-\sqrt{24})+2\cdot (5+\sqrt{24})}{(5+\sqrt{24})(5-\sqrt{24})}=\frac{20}{25-24}=20\\\\2)\; \; sina=-\frac{2}{5}\; ,\; \; cosa=-\frac{\sqrt{24}}{5}\\\\Q=\Big |\frac{-2/5}{1-\frac{\sqrt{24}}{5}}\Big |+\Big |\frac{-2/5}{1+\frac{\sqrt{24}}{5}}\Big |=20\\\\Otvet:\; \; 20\; .$