774 Преобразуйте выражение: a) ctg B-cosß-1 ; sinß б) 1 sina-1 1+ sina B) 1- ctgy tgy-1

Упражнение 774

Преобразуйте выражение:

а)

\(\operatorname{ctg} \beta - \frac{\cos \beta - 1}{\sin \beta} = \frac{\cos \beta}{\sin \beta} - \frac{\cos \beta - 1}{\sin \beta} = \frac{\cos \beta - \cos \beta + 1}{\sin \beta} = \frac{1}{\sin \beta}\)

Ответ: \(\frac{1}{\sin \beta}\)

б)

\(\frac{1}{\sin \alpha - 1} + \frac{1}{1 + \sin \alpha} = \frac{1 + \sin \alpha + \sin \alpha - 1}{(\sin \alpha - 1)(1 + \sin \alpha)} = \frac{2 \sin \alpha}{\sin^2 \alpha - 1} = \frac{2 \sin \alpha}{-\cos^2 \alpha} = -2 \sin \alpha \cdot \frac{1}{\cos^2 \alpha}\)

Ответ: \(\frac{-2 \sin \alpha}{\cos^2 \alpha}\)

в)

\(1 - \frac{\operatorname{ctg} \gamma}{\operatorname{tg} \gamma - 1} = \frac{\operatorname{tg} \gamma - 1 - \operatorname{ctg} \gamma}{\operatorname{tg} \gamma - 1} = \frac{\frac{\sin \gamma}{\cos \gamma} - 1 - \frac{\cos \gamma}{\sin \gamma}}{\frac{\sin \gamma}{\cos \gamma} - 1} = \frac{\sin^2 \gamma - \sin \gamma \cos \gamma - \cos^2 \gamma}{\sin \gamma \cos \gamma} : \frac{\sin \gamma - \cos \gamma}{\cos \gamma} =\)

\( = \frac{-(\cos^2 \gamma - \sin^2 \gamma) - \sin \gamma \cos \gamma}{\sin \gamma \cos \gamma} \cdot \frac{\cos \gamma}{\sin \gamma - \cos \gamma} = \frac{-(\cos \gamma - \sin \gamma)(\cos \gamma + \sin \gamma) - \sin \gamma \cos \gamma}{\sin \gamma} \cdot \frac{1}{\sin \gamma - \cos \gamma} =\)

\(= \frac{-(\cos \gamma + \sin \gamma) + \frac{\sin \gamma \cos \gamma}{\cos \gamma - \sin \gamma}}{\sin \gamma} = \frac{-\cos \gamma - \sin \gamma + \frac{\sin \gamma \cos \gamma}{\cos \gamma - \sin \gamma}}{\sin \gamma}\)

Ответ: \(\frac{-\cos \gamma - \sin \gamma + \frac{\sin \gamma \cos \gamma}{\cos \gamma - \sin \gamma}}{\sin \gamma}\)