Найдите значение выражения 1-cos2 x +1/sin2x,если sinx=-1/корень из 5

$sinx=-\frac{1}{\sqrt5}\\\\sinx<0\; \; \Rightarrow \; \; \; \pi <x<2\pi \\\\sin^2x+cos^2x=1\; \; \Rightarrow \; \; cosx=\pm \sqrt{1-sin^2x}\\\\cosx=\pm \sqrt{1-\frac{1}{5}}=\pm \sqrt{\frac{4}{5}}=\pm \frac{2}{\sqrt5}\\\\\underbrace {1-cos2x}_{2sin^2x}+\frac{1}{ain2x}=2sin^2x+\frac{1}{2\, sinx\, cosx}=A\\\\a)\; \; \pi <x<\frac{3\pi }{2}\, :\; cosx<0\; \to \; \; A=2\cdot (-\frac{1}{\sqrt5})^2+\frac{1}{2\cdot (-1/\sqrt5)\cdot (-2/\sqrt5)}=\\\\=\frac{2}{5}+\frac{5}{4}=\frac{8+25}{20}=\frac{33}{20}$

0\; \to \; A=2\cdot (-\frac{1}{\sqrt5})^2+\frac{1}{2\cdot (-1/\sqrt5)\cdot (2/\sqrt5)}=\\\\=10-\frac{5}{4}=\frac{2}{5}-\frac{5}{4}=-\frac{17}{20}" alt="b)\; \; \frac{3\pi }{2}0\; \to \; A=2\cdot (-\frac{1}{\sqrt5})^2+\frac{1}{2\cdot (-1/\sqrt5)\cdot (2/\sqrt5)}=\\\\=10-\frac{5}{4}=\frac{2}{5}-\frac{5}{4}=-\frac{17}{20}" align="absmiddle" class="latex-formula">