Avem
[tex]tg \ x = \dfrac{\sin x}{\cos x}; \ \ ctg \ x = \dfrac{\cos x}{\sin x}[/tex]
Atunci:
[tex]tg^2 \ x + ctg^2 \ x + 2 = \bigg(\dfrac{\sin x}{\cos x}\bigg)^2 + \bigg(\dfrac{\cos x}{\sin x} \bigg)^2 + 2 = \dfrac{\sin^2 x}{\cos^2 x} + \dfrac{\cos^2 x}{\sin^2 x} + 2\\[/tex]
[tex]= \dfrac{(\sin^2 x)^2 + (\cos^2 x)^2}{\sin^2 x \cdot \cos^2 x} + 2 = \dfrac{(\sin^2 x)^2 + (\cos^2 x)^2 + 2\sin^2 x \cdot \cos^2 x}{\sin^2 x \cdot \cos^2 x}\\[/tex]
[tex]= \dfrac{(\sin^2 x + \cos^2 x)^2}{\sin^2 x \cdot \cos^2 x} = \bigg(\dfrac{1}{\sin x \cdot \cos x}\bigg)^2[/tex]
unde:
[tex]\boxed{\boldsymbol{ \sin^{2} \alpha + \cos^{2} \alpha = 1 }}[/tex]
______
Alte teme rezolvate la trigonometrie:
- https://brainly.ro/tema/11057940
- https://brainly.ro/tema/11058857
- https://brainly.ro/tema/10587511
