Question

Zadanie dowodowe z trygonometrii w załączniku. Proszę o pomoc. ​

Answer 1

Odpowiedź:

Równość jest falszywa.

Szczegółowe wyjaśnienie:

[tex]\alpha+\beta+\gamma=\pi[/tex]

[tex]\alpha+\beta=\pi-\gamma[/tex]

[tex]\frac{tg\alpha+tg\beta}{tg\beta\cdot tg\gamma}=\frac{sin\alpha}{sin\beta \cdot sin\gamma}[/tex]

[tex]L=\frac{tg\alpha+tg\beta}{tg\beta\cdot tg\gamma}=\frac{\frac{sin\alpha}{cos\alpha}+\frac{sin\beta}{cos\beta}}{\frac{sin\beta}{cos\beta}\cdot\frac{sin\gamma}{cos\gamma}}=[/tex]

[tex]\frac{\frac{sin\alpha cos\beta}{cos\alpha cos\beta}+\frac{sin\beta cos\alpha}{cos\beta cos\alpha}}{\frac{sin\beta sin\gamma}{cos\beta cos\gamma}}=[/tex]

[tex]\frac{\frac{sin\alpha cos\beta+sin\beta cos\alpha}{cos\beta cos\alpha}}{\frac{sin\beta sin\gamma}{cos\beta cos\gamma}}=[/tex]

[tex]\frac{sin\alpha cos\beta+sin\beta cos\alpha}{cos\beta cos\alpha}:\frac{sin\beta sin\gamma}{cos\beta cos\gamma}=[/tex]

[tex]\frac{sin\alpha cos\beta+sin\beta cos\alpha}{cos\beta cos\alpha}\cdot\frac{cos\beta cos\gamma}{sin\beta sin\gamma}=[/tex]

[tex]\frac{sin\alpha cos\beta+cos\alpha sin\beta }{cos\alpha}\cdot\frac{cos\gamma}{sin\beta sin\gamma}=[/tex]

[tex]\frac{sin(\alpha +\beta) }{cos\alpha}\cdot\frac{cos\gamma}{sin\beta sin\gamma}=[/tex]

[tex]\frac{sin(\pi-\gamma) }{cos\alpha}\cdot\frac{cos\gamma}{sin\beta sin\gamma}=[/tex]

[tex]\frac{sin\gamma }{cos\alpha}\cdot\frac{cos\gamma}{sin\beta sin\gamma}=[/tex]

[tex]\frac{cos\gamma}{cos\alpha sin\beta }[/tex]

[tex]L\neq P[/tex]