[tex]\displaystyle\bf\\1)\\\\(\underbrace{1-Sin^{2} \alpha}_{Cos^{2} \alpha } )(\underbrace{1+tg^{2} \alpha }_{\dfrac{1}{Cos^{2} \alpha }} )=Cos^{2} \alpha \cdot \frac{1}{Cos^{2}\alpha } =1\\\\\\1=1 \ dokazano\\\\\\2)\\\\Sin^{2} \alpha (\underbrace{1+Ctg^{2} \alpha}_{ \dfrac{1}{Sin^{2} \alpha }} )-Cos^{2}\alpha = Sin^{2} \alpha \cdot\frac{1}{Sin^{2} \alpha } -Cos^{2}\alpha =\\\\\\=1-Cos^{2} \alpha =Sin^{2}\alpha \\\\\\Sin^{2} \alpha =Sin^{2} \alpha \ - \ dokazano[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
[tex]\displaystyle\bf\\1)\\\\(\underbrace{1-Sin^{2} \alpha}_{Cos^{2} \alpha } )(\underbrace{1+tg^{2} \alpha }_{\dfrac{1}{Cos^{2} \alpha }} )=Cos^{2} \alpha \cdot \frac{1}{Cos^{2}\alpha } =1\\\\\\1=1 \ dokazano\\\\\\2)\\\\Sin^{2} \alpha (\underbrace{1+Ctg^{2} \alpha}_{ \dfrac{1}{Sin^{2} \alpha }} )-Cos^{2}\alpha = Sin^{2} \alpha \cdot\frac{1}{Sin^{2} \alpha } -Cos^{2}\alpha =\\\\\\=1-Cos^{2} \alpha =Sin^{2}\alpha \\\\\\Sin^{2} \alpha =Sin^{2} \alpha \ - \ dokazano[/tex]