$\text{Dla}\alpha \in ⟨0^{\circ },{180}^{\circ }⟩:$   ${\sin }^2\alpha +{\cos }^2\alpha =1-\text{jedynka trygonometryczna}$   $\text{tg}\alpha =\frac{\sin \alpha }{\cos \alpha },\text{gdzie}\alpha \ne {90}^{\circ }$

---

Kąt α jest ostry, więc:

$\sin \alpha >0,\cos \alpha >0,\text{tg}\alpha >0,\text{ctg}\alpha >0$  

---

$\text{a)}{\sin }^2\alpha -1+{\cos }^2\alpha =\underset{1}{\underbrace{{\sin }^2\alpha +{\cos }^2\alpha }}-1=1-1=0$

---

$\text{b)}\frac{1-{\cos }^2\alpha }{1-{\sin }^2\alpha }=\frac{\stackrel{1}{\overbrace{{\sin }^2\alpha +{\cos }^2\alpha }}-{\cos }^2\alpha }{\underset{1}{\underbrace{{\sin }^2\alpha +{\cos }^2\alpha }}-{\sin }^2\alpha }=\frac{{\sin }^2\alpha }{{\cos }^2\alpha }={\left(\frac{\sin \alpha }{\cos \alpha }\right)}^2={\text{tg}}^2\alpha$  

---

$\text{c)}{\cos }^2\alpha \cdot \frac{{\sin }^2\alpha }{{\cos }^2\alpha }+{\cos }^2\alpha ={\sin }^2\alpha +{\cos }^2\alpha =1$  

---

$\text{d)}{\sin }^4\alpha +2{\sin }^2\alpha {\cos }^2\alpha +{\cos }^4\alpha ={\left({\sin }^2\alpha \right)}^2+2\cdot {\sin }^2\alpha \cdot {\cos }^2\alpha +{\left({\cos }^2\alpha \right)}^2=$  

Korzystamy ze wzoru skróconego mnożenia na kwadrat sumy.

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