569. Упростить выражение: 1) \(\frac{\operatorname{ctg}(\frac\pi2-\alpha)-\operatorname{tg}(\pi+\alpha)+\sin(\frac{3\pi}2-\alpha)}{\cos(\pi+\alpha)}\); 2) \(\frac{\sin(\pi-\alpha)+\cos(\frac\pi2+\alpha)+\operatorname{ctg}(\pi-\alpha)}{\operatorname{tg}(\frac{3\pi}2-\alpha)}\).

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Вопрос:

Accepted Answer

Применим формулы приведения:
\(\operatorname{ctg}(\frac\pi2-\alpha)=\operatorname{tg}\alpha\), \(\operatorname{tg}(\pi+\alpha)=\operatorname{tg}\alpha\), \(\sin(\frac{3\pi}2-\alpha)=-\cos\alpha\), \(\cos(\pi+\alpha)=-\cos\alpha\).
Тогда \(\frac{\operatorname{tg}\alpha-\operatorname{tg}\alpha-\cos\alpha}{-\cos\alpha}=1\).
\(\sin(\pi-\alpha)=\sin\alpha\), \(\cos(\frac\pi2+\alpha)=-\sin\alpha\), \(\operatorname{ctg}(\pi-\alpha)=-\operatorname{ctg}\alpha\), а \(\operatorname{tg}(\frac{3\pi}2-\alpha)=\operatorname{ctg}\alpha\).
Следовательно, \(\frac{\sin\alpha-\sin\alpha-\operatorname{ctg}\alpha}{\operatorname{ctg}\alpha}=-1\).

Ответ: 1) 1; 2) −1.