Zadanie 2
(tg\alpha+ctg\alpha):(\frac{1}{\sin\alpha}-\frac{1}{cos\alpha})=\frac{1}{cos\alpha-sin\alpha}
L=(tg\alpha+ctg\alpha):(\frac{1}{\sin\alpha}-\frac{1}{cos\alpha})=
=(\frac{sin\alpha}{cos\alpha}+\frac{cos\alpha}{sin\alpha}):\frac{cos\alpha-sin\alpha}{sin\alpha*cos\alpha}=
=\frac{sin\alpha*sin\alpha+cos\alpha*cos\alpha}{sin\alpha*cos\alpha}*\frac{sin\alpha*cos\alpha}{cos\alpha-sin\alpha}=
=\frac{sin^2\alpha+cos^2\alpha}{cos\alpha-sin\alpha}=
=\frac{1}{cos\alpha-sin\alpha}
L=P