Podstawowe tożsamości trygonometryczne:

$\text{1)}{\sin }^2\alpha +{\cos }^2\alpha =1,\text{dla}\alpha \in ⟨0^{\circ },{180}^{\circ }⟩$  

$\text{2)}\text{tg}\alpha \cdot \text{ctg}\alpha =1,\text{dla}\alpha \in \left(0^{\circ },{90}^{\circ }\right)\cup \left({90}^{\circ },{180}^{\circ }\right)$  

$\text{3)}\frac{\sin \alpha }{\cos \alpha }=\text{tg}\alpha ,\text{dla}\alpha \in ⟨0^{\circ },{90}^{\circ })\cup ({90}^{\circ },{180}^{\circ }⟩$  

$\text{4)}\frac{\cos \alpha }{\sin \alpha }=\text{ctg}\alpha ,\text{dla}\alpha \in \left(0^{\circ },{180}^{\circ }\right)$  

Tożsamość 1) nazywamy jedynką trygonometryczną.

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$\text{a)}$  

Z jedynki trygonometrycznej mamy:

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

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

 

$\frac{{\sin }^2\alpha }{1-{\sin }^2\alpha }=\frac{{\sin }^2\alpha }{{\cos }^2\alpha }={\mathrm{tg}}^2\alpha ={\left(\frac{5}{2}\right)}^2=\frac{25}{4}=6\frac{1}{4}$  

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$\text{b)}$  

Z treści zadania wiemy, że:

$\mathrm{tg}\alpha =2\frac{1}{2}$  

$\mathrm{tg}\alpha =\frac{5}{2}$  

$\frac{\sin \alpha }{\cos \alpha }=\frac{5}{2}$  

$\sin \alpha =\frac{5}{2}\cos \alpha$  

 

$\frac{\sin \alpha -\cos \alpha }{\cos \alpha }=\frac{\frac{5}{2}\cos \alpha -\cos \alpha }{\cos \alpha }=$  

$=\frac{\frac{5}{2}\cos \alpha -\frac{2}{2}\cos \alpha }{\cos \alpha }=\frac{\frac{3}{2}\cos \alpha }{\cos \alpha }=\frac{\frac{3}{2}}{1}=\frac{3}{2}$