sprawdź tożsamość

zuza52178 · Post autor: **zuza52178** » 31 mar 2011, o 11:36

\(\displaystyle{ \frac{1}{(1+\sin \alpha )(1-\sin \alpha )} = \tg ^{2} \alpha +1}\)

Errichto · Post autor: **Errichto** » 31 mar 2011, o 11:40

\(\displaystyle{ L= \frac{1}{1- \sin^2 \alpha }= \frac{1}{ \cos^2 \alpha }= \frac{ \sin^2 \alpha + \cos^2 \alpha }{ \cos^2 \alpha}= \frac{ \sin^2 \alpha }{ \cos^2 \alpha }+ \frac{ \cos^2 \alpha }{ \cos^2 \alpha }= \tg^2 \alpha +1 =P}\)