a)

$\frac{1}{1-\cos \alpha }+\frac{1}{1+\cos \alpha }=\frac{2}{{\sin }^2\alpha }$  

 

Założenia:

$1-\cos \alpha \ne 0$  

$\cos \alpha \ne 1$  

$\alpha \ne 0^{\circ }$       

 

$1+\cos \alpha \ne 0$  

$\cos \alpha \ne -1$  

$\alpha \ne {180}^{\circ }$  

 

${\sin }^2\alpha \ne 0$  

$\alpha \ne 0^{\circ }$  

$\alpha \ne {180}^{\circ }$   

 

$L=\frac{1}{1-\cos \alpha }+\frac{1}{1+\cos \alpha }=\frac{1+\cos \alpha }{\left(1+\cos \alpha \right)\cdot \left(1-\cos \alpha \right)}+\frac{1-\cos \alpha }{\left(1+\cos \alpha \right)\left(1-\cos \alpha \right)}=\frac{1+\cos \alpha +1-\cos \alpha }{\left(1+\cos \alpha \right)\left(1-\cos \alpha \right)}=\frac{2}{1-{\cos }^2\alpha }=\frac{2}{{\sin }^2\alpha }$  

 

 

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b)

$\frac{1+\cos \alpha }{1-\cos \alpha }-\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{4}{\sin \alpha \cdot \mathrm{tg}\alpha }$  

 

Założenia:

$1-\cos \alpha \ne 0$  

$\cos \alpha \ne 1$  

$\alpha \ne 0^{\circ }$  

 

$1+\cos \alpha \ne 0$  

$\cos \alpha \ne -1$  

$\alpha \ne {180}^{\circ }$  

           

$\sin \alpha \ne 0$  

$\alpha \ne 0^{\circ }$  

$\alpha \ne {180}^{\circ }$  

 

$\mathrm{tg}\alpha \ne 0$  

$\alpha \ne 0^{\circ }$  

 

$L=\frac{1+\cos \alpha }{1-\cos \alpha }-\frac{1-\cos \alpha }{1+\cos \alpha }=$   (sprowadzamy ułamki do wspólnego mianownika)

 

$=\frac{\left(1+\cos \alpha \right)\cdot \left(1+\cos \alpha \right)}{\left(1+\cos \alpha \right)\left(1-\cos \alpha \right)}-\frac{\left(1-\cos \alpha \right)\left(1-\cos \alpha \right)}{\left(1+\cos \alpha \right)\left(1-\cos \alpha \right)}=$  (korzystamy ze wzorów skróconego mnożenia)

 

$=\frac{1+2\cos \alpha +{\cos }^2\alpha }{1-{\cos }^2}-\frac{1-2\cos \alpha +{\cos }^2\alpha }{1-{\cos }^2}=$  (w mianowniku korzystajmy z jedynki trygonometrycznej)

 

$=\frac{1+2\cos \alpha +{\cos }^2\alpha -\left(1-2\cos \alpha +{\cos }^2\alpha \right)}{{\sin }^2\alpha }=$   

 

$=\frac{1+2\cos \alpha +{\cos }^2\alpha -1+2\cos \alpha -{\cos }^2\alpha }{{\sin }^2\alpha }=$   

 

$=\frac{4\cos \alpha }{{\sin }^2\alpha }=$  

 

$=\frac{4}{\frac{\sin \alpha \cdot \sin \alpha }{\cos \alpha }}=$  

 

$=\frac{4}{\sin \alpha \cdot \mathrm{tg}\alpha }=P$  

 

 

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  c)

$\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }=\frac{1}{{\cos }^2\alpha \cdot \mathrm{tg}\alpha }$  

 

Założenia:

$\cos \alpha \ne 0$  

$\alpha \ne {90}^{\circ }$  

 

$\sin \alpha \ne 0^{\circ }$  

$\alpha \ne 0^{\circ }$  

$\alpha \ne {180}^{\circ }$        

         

$\mathrm{tg}\alpha \ne 0^{\circ }$  

$\alpha \ne 0^{\circ }$          

 

   

$L=\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }=$    (sprowadzamy do wspólnego mianownika)

 

$=\frac{\sin \alpha \cdot \sin \alpha }{\sin \alpha \cdot \cos \alpha }+\frac{\cos \alpha \cdot \cos \alpha }{\sin \alpha \cdot \cos \alpha }=$   

 

$=\frac{{\sin }^2\alpha }{\sin \alpha \cdot \cos \alpha }+\frac{{\cos }^2\alpha }{\sin \alpha \cdot \cos \alpha }=$  

 

$=\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cdot \cos \alpha }=$   (korzystamy z jedynki trygonometrycznej)

 

$=\frac{1}{\sin \alpha \cdot \cos \alpha }$  

 

$P=\frac{1}{{\cos }^2\alpha \cdot \mathrm{tg}\alpha }=\frac{1}{{\cos }^2\alpha \cdot \frac{\sin \alpha }{\cos \alpha }}=\frac{1}{\cos \alpha \cdot \sin \alpha }=L$  

 

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d)

  $\frac{1}{\mathrm{tg}\alpha -1}+\frac{1}{\mathrm{tg}\alpha +1}=\frac{2\sin \alpha \cdot \cos \alpha }{\left(\sin \alpha +\cos \alpha \right)\cdot \left(\sin \alpha -\cos \alpha \right)}$  

 

Założenia:

$\mathrm{tg}\alpha -1\ne 0$  

$\mathrm{tg}\alpha \ne 1$  

$\alpha \ne {45}^{\circ }$  

 

$\mathrm{tg}\alpha +1\ne 0$  

$\mathrm{tg}\alpha \ne -1$  

$\alpha \ne {135}^{\circ }$  

 

$\sin \alpha +\cos \alpha \ne 0$  

$\sin \alpha \ne -\cos \alpha$            

$\alpha \ne {135}^{\circ }$   (zauważmy, że $\sin {135}^{\circ }=\sin {45}^{\circ }=\frac{\sqrt{2}}{2}$  , oraz  $\cos {135}^{\circ }=\cos \left({180}^{\circ }-{45}^{\circ }\right)=-\cos {45}^{\circ }=-\frac{\sqrt{2}}{2})$  

 

$\sin \alpha -\cos \alpha \ne 0$  

$\sin \alpha \ne \cos \alpha$  

$\alpha \ne {45}^{\circ }$  

 

 

$L=\frac{1}{\mathrm{tg}\alpha -1}+\frac{1}{\mathrm{tg}\alpha +1}=$  

 

$=\frac{1}{\frac{\sin \alpha }{\cos \alpha }-\frac{\cos \alpha }{\cos \alpha }}+\frac{1}{\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\cos \alpha }}=$  

 

$=\frac{1}{\frac{\sin \alpha -\cos \alpha }{\cos \alpha }}+\frac{1}{\frac{\sin \alpha +\cos \alpha }{\cos \alpha }}=$  

 

$=\frac{\cos \alpha }{\sin \alpha -\cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }=$   (sprowadźmy ułamki do wspólnego mianownika)

 

$=\frac{\cos \alpha \left(\sin \alpha +\cos \alpha \right)}{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}+\frac{\cos \alpha \left(\sin \alpha -\cos \alpha \right)}{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}=$  

 

$=\frac{\cos \alpha \cdot \sin \alpha +{\cos }^2\alpha }{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}+\frac{\cos \alpha \cdot \sin \alpha -{\cos }^2\alpha }{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}=$  

 

$=\frac{\cos \alpha \cdot \sin \alpha +{\cos }^2\alpha +\cos \alpha \cdot \sin \alpha -{\cos }^2\alpha }{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}=$  

 

$=\frac{\cos \alpha \cdot \sin \alpha +\cos \alpha \cdot \sin \alpha }{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}=$              

 

$=\frac{2\cdot \sin \alpha \cdot \cos \alpha }{\left(\sin \alpha -\cos \alpha \right)\left(\sin \alpha +\cos \alpha \right)}=P$