1)
[tex]\it sin^25^o+sin^245^o+sin^285^o =sin^25^o+\Big(\dfrac{\sqrt2}{2}\Big)^2+cos^2(90^o-85^o) =\\ \\ \\ =sin^25^o+cos^25^o+\dfrac{\ 2^{(2}}{4}=1+\dfrac{1}{2} =\dfrac{3}{2}[/tex]
2)
[tex]\it ctg10^o -ctg80^o=\dfrac{cos10^o}{sin10^o}-\dfrac{cos80^o}{sin80^o}= \dfrac{cos10^o}{sin10^o}-\dfrac{sin10^o}{cos10^o}=\\ \\ \\ =\dfrac{cos^210^o-sin^210^0}{sin10^ocos10^o}=\dfrac{cos20^o}{\dfrac{1}{2}sin20^o}=\dfrac{2cos20^o}{sin20^0}\ \ \ \ \ (1)\\ \\ \\ cos70^o=sin20^o\ \ \ \ \ (2)[/tex]
Folosind relațiile (1), (2), membrul stâng al egalității devine:
[tex]\it \dfrac{2cos20^o}{sin20^o}\cdot sin20^o=2cos20^o=2sin70^o=2sin(180^o-70^o)=2sin110^o[/tex]
