Registado: 25 mar 2012, 19:59 Mensagens: 1026 Localização: Rio de Janeiro - Brasil Agradeceu: 116 vezes Foi agradecido: 204 vezes
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Olá! Seja bem-vindo diegorodrigo2022!!
\(\mathbf{U = \frac{(1 - \sin \alpha + \cos \alpha)^2}{(\sin \alpha + \tan \alpha)(\cos \alpha - \cot \alpha)}}\)
\(\mathbf{U = \frac{(1 - \sin \alpha)^2 + 2 \cdot (1 - \sin \alpha) \cdot \cos \alpha + \cos^2 \alpha}{\left ( \sin \alpha + \frac{\sin \alpha}{\cos \alpha} \right )\left ( \cos \alpha - \frac{\cos \alpha}{\sin \alpha} \right )}}\)
\(\displaystyle \mathbf{U = \frac{1 - 2 \cdot \sin \alpha + \sin^2 \alpha + 2 \cdot \cos \alpha - 2 \cdot \sin \alpha \cdot \cos \alpha + \cos^2 \alpha}{\frac{\sin \alpha \cdot \cos \alpha + \sin \alpha}{\cos \alpha} \cdot \frac{\sin \alpha \cdot \cos \alpha - \cos \alpha}{sin \alpha}}}\)
\(\displaystyle \mathbf{U = \frac{2 - 2 \cdot \sin \alpha + 2 \cdot \cos \alpha - 2 \cdot \sin \alpha \cdot \cos \alpha}{\frac{\sin \alpha \cdot \cos \alpha + \sin \alpha}{\cos \alpha} \cdot \frac{\sin \alpha \cdot \cos \alpha - \cos \alpha}{sin \alpha}}}\)
\(\displaystyle \mathbf{U = \frac{2 \cdot (1 - \sin \alpha) + 2 \cdot \cos \alpha \cdot (1 - \sin \alpha)}{\frac{\sin \alpha \cdot (\cos \alpha + 1)}{\cos \alpha} \cdot \frac{\cos \alpha \cdot (\sin \alpha - 1)}{sin \alpha}}}\)
\(\mathbf{U = \frac{(1 - \sin \alpha) \cdot 2 \cdot (1 + \cos \alpha)}{(\cos \alpha + 1) \cdot (\sin \alpha - 1)}}\)
\(\mathbf{U = \frac{(1 - \sin \alpha) \cdot 2}{(\sin \alpha - 1)}}\)
\(\mathbf{U = \frac{- 1 \cdot 2}{1}}\)
\(\fbox{\fbox{\mathbf{U = - 2}}}\)
_________________ Daniel Ferreirase gosta da resposta, RESPONDA A QUEM PRECISA
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