Zamiast tgα, zastosuję angielskie oznaczenie tanα.
[tex]\tan\alpha=\dfrac{1}{7}\\\\\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{1}{7}\\\\\cos\alpha=7\sin\alpha\\\\\dfrac{\cos\alpha-2\sin\alpha}{2\cos\alpha+\sin\alpha}=\dfrac{7\sin\alpha-2\sin\alpha}{2\cdot7\sin\alpha+\sin\alpha}=\dfrac{5\sin\alpha}{14\sin\alpha+\sin\alpha}=\dfrac{5\sin\alpha}{15\sin\alpha}=\dfrac{1}{3}[/tex]
