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Câu hỏi: Chứng minh rằng:
Giải chi tiết:
\(\begin{array}{l}\dfrac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha - {{\cos }^2}\alpha }} = \dfrac{{{{\sin }^2}\alpha \left( {\dfrac{1}{{{{\cos }^2}\alpha }} - 1} \right)}}{{{{\cos }^2}\alpha \left({\dfrac{1}{{{{\sin }^2}\alpha }} - 1} \right)}}\\ = \dfrac{{{{\sin }^2}\alpha {{\tan }^2}\alpha }}{{{{\cos }^2}\alpha {{\cot }^2}\alpha }} = {\tan ^6}\alpha \end{array}\)
Giải chi tiết:
\(\begin{array}{l}\dfrac{{\sin \alpha + \cos \alpha }}{{{{\cos }^3}\alpha }} = \dfrac{{\cos \alpha \left( {\tan \alpha + 1} \right)}}{{{{\cos }^3}\alpha }}\\ = \left({{{\tan }^2}\alpha + 1} \right)\left({\tan \alpha + 1} \right)\\ = 1 + \tan \alpha + {\tan ^2}\alpha + {\tan ^3}\alpha .\end{array}\)
Giải chi tiết:
\(\begin{array}{l}\sqrt {{{\sin }^2}\alpha \left( {1 + \cot \alpha } \right) + {{\cos }^2}\alpha \left({1 + \tan \alpha } \right)} \\ = \sqrt {{{\sin }^2}\alpha + \sin \alpha \cos \alpha + {{\cos }^2}\alpha + \cos \alpha \sin \alpha } \\ = \sqrt {{{\left({\sin \alpha + \cos \alpha } \right)}^2}} = \left| {\sin \alpha + \cos \alpha } \right|.\end{array}\)
Giải chi tiết:
\(\begin{array}{l}{\sin ^2}\alpha {\tan ^2}\alpha + 4{\sin ^2}\alpha - {\tan ^2}\alpha + 3{\cos ^2}\alpha \\ = - {\tan ^2}\alpha {\cos ^2}\alpha + 4{\sin ^2}\alpha + 3{\cos ^2}\alpha \\ = 3\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = 3\end{array}\)
Câu a
\(\dfrac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha - {{\cos }^2}\alpha }} = {\tan ^6}\alpha ;\)Giải chi tiết:
\(\begin{array}{l}\dfrac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha - {{\cos }^2}\alpha }} = \dfrac{{{{\sin }^2}\alpha \left( {\dfrac{1}{{{{\cos }^2}\alpha }} - 1} \right)}}{{{{\cos }^2}\alpha \left({\dfrac{1}{{{{\sin }^2}\alpha }} - 1} \right)}}\\ = \dfrac{{{{\sin }^2}\alpha {{\tan }^2}\alpha }}{{{{\cos }^2}\alpha {{\cot }^2}\alpha }} = {\tan ^6}\alpha \end{array}\)
Câu b
\(\dfrac{{\sin \alpha + \cos \alpha }}{{{{\cos }^3}\alpha }} = 1 + \tan \alpha + {\tan ^2}\alpha + {\tan ^3}\alpha ;\)Giải chi tiết:
\(\begin{array}{l}\dfrac{{\sin \alpha + \cos \alpha }}{{{{\cos }^3}\alpha }} = \dfrac{{\cos \alpha \left( {\tan \alpha + 1} \right)}}{{{{\cos }^3}\alpha }}\\ = \left({{{\tan }^2}\alpha + 1} \right)\left({\tan \alpha + 1} \right)\\ = 1 + \tan \alpha + {\tan ^2}\alpha + {\tan ^3}\alpha .\end{array}\)
Câu c
\(\sqrt {{{\sin }^2}\alpha \left( {1 + \cot \alpha } \right) + {{\cos }^2}\alpha \left({1 + \tan \alpha } \right)} = \left| {\sin \alpha + \cos \alpha } \right|\)Giải chi tiết:
\(\begin{array}{l}\sqrt {{{\sin }^2}\alpha \left( {1 + \cot \alpha } \right) + {{\cos }^2}\alpha \left({1 + \tan \alpha } \right)} \\ = \sqrt {{{\sin }^2}\alpha + \sin \alpha \cos \alpha + {{\cos }^2}\alpha + \cos \alpha \sin \alpha } \\ = \sqrt {{{\left({\sin \alpha + \cos \alpha } \right)}^2}} = \left| {\sin \alpha + \cos \alpha } \right|.\end{array}\)
Câu d
\({\sin ^2}\alpha {\tan ^2}\alpha + 4{\sin ^2}\alpha - {\tan ^2}\alpha + 3{\cos ^2}\alpha = 3\).Giải chi tiết:
\(\begin{array}{l}{\sin ^2}\alpha {\tan ^2}\alpha + 4{\sin ^2}\alpha - {\tan ^2}\alpha + 3{\cos ^2}\alpha \\ = - {\tan ^2}\alpha {\cos ^2}\alpha + 4{\sin ^2}\alpha + 3{\cos ^2}\alpha \\ = 3\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = 3\end{array}\)
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