- LG a
- LG b
- LG c
- LG d
Chứng minh rằng:
LG a
\[\dfrac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha - {{\cos }^2}\alpha }} = {\tan ^6}\alpha ;\]
Lời giải chi tiết:
\[\begin{array}{l}\dfrac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha - {{\cos }^2}\alpha }} = \dfrac{{{{\sin }^2}\alpha \left[ {\dfrac{1}{{{{\cos }^2}\alpha }} - 1} \right]}}{{{{\cos }^2}\alpha \left[ {\dfrac{1}{{{{\sin }^2}\alpha }} - 1} \right]}}\\ = \dfrac{{{{\sin }^2}\alpha {{\tan }^2}\alpha }}{{{{\cos }^2}\alpha {{\cot }^2}\alpha }} = {\tan ^6}\alpha \end{array}\]
LG b
\[\dfrac{{\sin \alpha + \cos \alpha }}{{{{\cos }^3}\alpha }} = 1 + \tan \alpha + {\tan ^2}\alpha + {\tan ^3}\alpha ;\]
Lời giải chi tiết:
\[\begin{array}{l}\dfrac{{\sin \alpha + \cos \alpha }}{{{{\cos }^3}\alpha }} = \dfrac{{\cos \alpha \left[ {\tan \alpha + 1} \right]}}{{{{\cos }^3}\alpha }}\\ = \left[ {{{\tan }^2}\alpha + 1} \right]\left[ {\tan \alpha + 1} \right]\\ = 1 + \tan \alpha + {\tan ^2}\alpha + {\tan ^3}\alpha .\end{array}\]
LG c
\[\sqrt {{{\sin }^2}\alpha \left[ {1 + \cot \alpha } \right] + {{\cos }^2}\alpha \left[ {1 + \tan \alpha } \right]} = \left| {\sin \alpha + \cos \alpha } \right|\]
Lời giải chi tiết:
\[\begin{array}{l}\sqrt {{{\sin }^2}\alpha \left[ {1 + \cot \alpha } \right] + {{\cos }^2}\alpha \left[ {1 + \tan \alpha } \right]} \\ = \sqrt {{{\sin }^2}\alpha + \sin \alpha \cos \alpha + {{\cos }^2}\alpha + \cos \alpha \sin \alpha } \\ = \sqrt {{{\left[ {\sin \alpha + \cos \alpha } \right]}^2}} = \left| {\sin \alpha + \cos \alpha } \right|.\end{array}\]
LG d
\[{\sin ^2}\alpha {\tan ^2}\alpha + 4{\sin ^2}\alpha - {\tan ^2}\alpha + 3{\cos ^2}\alpha = 3\].
Lời giải chi tiết:
\[\begin{array}{l}{\sin ^2}\alpha {\tan ^2}\alpha + 4{\sin ^2}\alpha - {\tan ^2}\alpha + 3{\cos ^2}\alpha \\ = - {\tan ^2}\alpha {\cos ^2}\alpha + 4{\sin ^2}\alpha + 3{\cos ^2}\alpha \\ = 3\left[ {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right] = 3\end{array}\]