December 2018 1 82 Report
Wykaż, że dla kąta ostrego α tożsamością jest równość:

a)  1+ctg^{2} \alpha = \frac{1}{sin ^{2} \alpha}
b)  \frac{sin \alpha }{1+cos \alpha } - \frac{1-cos \alpha }{sin \alpha } =0
c)  \frac{1}{tg \alpha +ctg \alpha } =sin \alpha cos \alpha
d)  \frac{ sin^{2} \alpha }{1-cos \alpha } =1+cos \alpha

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