Страница 151 номер 511, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество \(\dfrac{\sin^2 \alpha}{\cos \alpha (1 + \operatorname{ctg} \alpha)} - \dfrac{\cos^2 \alpha}{\sin \alpha (1 + \operatorname{tg} \alpha)} = \dfrac{2\sqrt{2} \sin\left(\alpha - \frac{\pi}{4}\right)}{\sin 2\alpha}.\)
\(\dfrac{\sin^{2} \alpha}{\cos \alpha(1 + \operatorname{ctg} \alpha)} - \dfrac{\cos^{2} \alpha}{\sin \alpha(1 + \operatorname{tg} \alpha)} = \dfrac{2\sqrt{2}\sin\left(\alpha - \frac{\pi}{4}\right)}{\sin 2\alpha}\)
\(\dfrac{\sin^{2} \alpha}{\cos \alpha(1 + \operatorname{ctg} \alpha)} - \dfrac{\cos^{2} \alpha}{\sin \alpha(1 + \operatorname{tg} \alpha)} =\)
\(= \dfrac{\sin^{3} \alpha}{\cos \alpha(\sin \alpha + \cos \alpha)} - \dfrac{\cos^{3} \alpha}{\sin \alpha(\cos \alpha + \sin \alpha)} =\)
\(= \dfrac{\sin^{4} \alpha - \cos^{4} \alpha}{\cos \alpha \sin \alpha(\sin \alpha + \cos \alpha)} =\)
\(= \dfrac{(\sin^{2} \alpha - \cos^{2} \alpha)(\sin^{2} \alpha + \cos^{2} \alpha)}{\cos \alpha \sin \alpha(\sin \alpha + \cos \alpha)} = \dfrac{\sin \alpha - \cos \alpha}{\cos \alpha \sin \alpha};\)
\(\dfrac{2\sqrt{2}\sin\left(\alpha - \frac{\pi}{4}\right)}{\sin 2\alpha} = \dfrac{2\sqrt{2}\left(\sin \alpha \cos\frac{\pi}{4} - \cos \alpha \sin\frac{\pi}{4}\right)}{2\sin \alpha \cos \alpha} =\)
\(= \dfrac{2\sqrt{2} \cdot \frac{\sqrt{2}}{2}\sin \alpha - 2\sqrt{2} \cdot \frac{\sqrt{2}}{2}\cos \alpha}{2\sin \alpha \cos \alpha} = \dfrac{\sin \alpha - \cos \alpha}{\cos \alpha \sin \alpha};\)
Левая и правая части тождества равны.
