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zadania w załączniku

Zadania W Załączniku class=

Odpowiedź :

[tex]zad.1\\\\cos\alpha \cdot tg\alpha - sin\alpha \cdot ctg\alpha =cos\alpha \cdot \dfrac{sin\alpha }{cos\alpha } -sin\alpha \cdot \dfrac{cos\alpha }{sin\alpha } =sin\alpha - cos\alpha \\\\cos\alpha \cdot tg\alpha - sin\alpha \cdot ctg\alpha =sin\alpha - cos\alpha~~~\land ~~\alpha =45^{o} ~~\Rightarrow ~~sin45^{o}-cos45^{o}=\dfrac{\sqrt{2} }{2} -\dfrac{\sqrt{2} }{2}=0\\\\\\Odp:~~A.~~0[/tex]

[tex]zad.2\\\\\dfrac{sin13^{o}\cdot cos^{2}33^{o}+cos77^{o} \cdot cos^{2}57^{o} }{cos77^{o} } =\dfrac{sin13^{o}\cdot cos^{2}33^{o} }{cos77^{o} }+\dfrac{cos77^{o} \cdot cos^{2}57^{o} }{cos77^{o} }=[/tex]

[tex]=\dfrac{sin(90^{o}-77^{o} ) \cdot cos^{2}33^{o} }{cos77^{o} }+cos^{2} 57^{o} =\dfrac{cos77^{o} \cdot cos^{2}33^{o} }{cos77^{o} }+cos^{2} 57^{o} =cos^{2} 33^{o} +cos^{2} 57^{o} =cos^{2} (90^{o} -57^{o} )+cos^{2} 57^{o} =sin^{2} 57^{o}+cos^{2} 57^{o}=1\\\\\\korzystam~~ze~~wzorow:\\\\sin^{2} \alpha +cos^{2} \alpha =1\\sin(90^{o} -\alpha )=cos\alpha \\\\cos(90^{o} -\alpha )=sin\alpha\\\\Odp:~~B.~~1[/tex]