Odpowiedź:
[tex]\huge\boxed{5+2\sqrt5}[/tex]
Szczegółowe wyjaśnienie:
[tex]\left(\dfrac{1}{\sin\alpha}+\dfrac{1}{\cos\alpha}\right)^2=\left(\dfrac{\cos\alpha}{\sin\alpha\cos\alpha}+\dfrac{\sin\alpha}{\sin\alpha\cos\alpha}\right)^2=\left(\dfrac{\cos\alpha+\sin\alpha}{\sin\alpha\cos\alpha}\right)^2[/tex]
skorzystamy z twierdzeń oraz wzoru skróconego mnożenia:
[tex]\left(\dfrac{a}{b}\right)^2=\dfrac{a^2}{b^2}\\\\(a\cdot b)^2=a^2\cdot b^2\\\\(a+b)^2=a^2+2ab+b^2[/tex]
[tex]=\dfrac{(\cos\alpha+\sin\alpha)^2}{(\sin\alpha\cos\alpha)^2}=\dfrac{\cos^2\alpha+2\sin\alpha\cos\alpha+\sin^2\alpha}{(\sin\alpha\cos\alpha)^2}\\\\=\dfrac{(\cos^2\alpha+\sin^2\alpha)+2\sin\alpha\cos\alpha}{(\sin\alpha\cos\alpha)^2}[/tex]
skorzystamy z jedynki trygonometrycznej [tex]\sin^2\alpha+\cos^2\alpha=1[/tex]:
[tex]=\dfrac{1+2\sin\alpha\cos\alpha}{(\sin\alpha\cos\alpha)^2}[/tex]
Podstawiamy [tex]\sin\alpha\cos\alpha=\dfrac{\sqrt5}{5}[/tex]
[tex]\left(1+2\cdot\dfrac{\sqrt5}{5}\right):\left(\dfrac{\sqrt5}{5}\right)^2=\left(1+\dfrac{2\sqrt5}{5}\right):\dfrac{5\!\!\!\!\diagup^1}{25\!\!\!\!\!\diagup_5}=\left(\dfrac{5}{5}+\dfrac{2\sqrt5}{5}\right)\cdot5\\\\=\dfrac{5+2\sqrt5}{5\!\!\!\!\diagup_1}\cdot5\!\!\!\!\diagup^1=5+2\sqrt5[/tex]
