Question

Tożsamości trygonometryczne
1-cos^2x/sinxcosx=tgx

Answer 1

Odpowiedź:

Korzystamy ze wzoru:

• [tex]sin^{2} \alpha +cos^{2} \alpha =1~~\Rightarrow~~\huge\boxed{cos^{2} \alpha =1-sin^{2} \alpha}[/tex]
• [tex]\huge\boxed{\dfrac{sin\alpha }{cos\alpha } =tg\alpha }[/tex]

[tex]\huge\boxed{\dfrac{1-cos^{2} \alpha }{sin\alpha \cdot cos\alpha } =tg\alpha }[/tex]

[tex]L=\dfrac{1-cos^{2} \alpha }{sin\alpha \cdot cos\alpha } =\dfrac{1-(1-sin^{2} \alpha )}{sin\alpha \cdot cos\alpha } =\dfrac{1-1+sin^{2} \alpha }{sin\alpha \cdot cos\alpha } =\dfrac{sin^{2}\alpha \!\!\!\!\!\diagup^s^i^n^\alpha }{sin\alpha\!\!\!\!\!\diagup_1 \cdot cos\alpha } =\dfrac{sin\alpha }{cos\alpha } =tg\alpha[/tex]

[tex]P=tg\alpha \\\\\\L=P~~~~cbdu[/tex]