Ответ:
- а) \( g(-\alpha)\cos\alpha+\sin\alpha=- g\alpha\cos\alpha+\sin\alpha=-\sin\alpha+\sin\alpha=0\).
- б) \(\dfrac{\ctg(-\alpha)\sin\alpha}{\cos\alpha}=\dfrac{-\ctg\alpha\sin\alpha}{\cos\alpha}=\dfrac{-\dfrac{\cos\alpha}{\sin\alpha}\sin\alpha}{\cos\alpha}=-1\).
- в) \(\cos^2\alpha g^2(-\alpha)-1=\cos^2\alpha g^2\alpha-1=\cos^2\alpha\cdot\dfrac{\sin^2\alpha}{\cos^2\alpha}-1=\sin^2\alpha-1=-\cos^2\alpha\).
- г) \(\dfrac{1- g(-\alpha)}{\sin\alpha+\cos(-\alpha)}=\dfrac{1+ g\alpha}{\sin\alpha+\cos\alpha}=\dfrac{1+\dfrac{\sin\alpha}{\cos\alpha}}{\sin\alpha+\cos\alpha}=\dfrac{\dfrac{\sin\alpha+\cos\alpha}{\cos\alpha}}{\sin\alpha+\cos\alpha}=\dfrac{1}{\cos\alpha}\).
Ответ: а) 0; б) −1; в) −cos² α; г) 1/cos α.
