Twierdzenie tangensów

Z twierdzenia sinusów mamy \[ \frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=2 R \] Zatem: \[a=2 R \sin \alpha\] \[b=2 R \sin \beta\] Zatem: \[ \frac{a-b}{a+b}=\frac{2 R \sin \alpha-2 R \sin \beta}{2 R \sin \alpha+2 R \sin \beta}=\frac{\sin \alpha-\sin \beta}{\sin \alpha+\sin \beta} . \] Teraz używamy wzorów na sumę i różnicę sinusów: \[ \begin{aligned} & \sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}, \\ & \sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} . \end{aligned} \] Czyli: \[ \frac{\sin \alpha-\sin \beta}{\sin \alpha+\sin \beta} =\frac{2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}} =\frac{\sin \frac{\alpha-\beta}{2}}{\cos \frac{\alpha-\beta}{2}} \cdot \frac{\cos \frac{\alpha+\beta}{2}}{\sin \frac{\alpha+\beta}{2}} . \] Ze wzoru \(\operatorname{tg} x=\frac{\sin x}{\cos x}\) mamy: \[\frac{\sin \frac{\alpha-\beta}{2}}{\cos \frac{\alpha-\beta}{2}}=\operatorname{tg} \frac{\alpha-\beta}{2}\] oraz: \[\frac{\sin \frac{\alpha+\beta}{2}}{\cos \frac{\alpha+\beta}{2}}=\operatorname{tg} \frac{\alpha+\beta}{2}\] Zatem: \[ \frac{a-b}{a+b}=\operatorname{tg} \frac{\alpha-\beta}{2} \cdot \frac{1}{\operatorname{tg} \frac{\alpha+\beta}{2}}=\frac{\operatorname{tg} \frac{\alpha-\beta}{2}}{\operatorname{tg} \frac{\alpha+\beta}{2}} . \] Co kończy dowód.