Câu 6.28 trang 199 SBT Đại số 10 Nâng cao

Chứng minh rằng:

a) \(\dfrac{{{{\tan }^2}\alpha  - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha  - {{\cos }^2}\alpha }} = {\tan ^6}\alpha ;\)

b) \(\dfrac{{\sin \alpha  + \cos \alpha }}{{{{\cos }^3}\alpha }} = 1 + \tan \alpha  + {\tan ^2}\alpha  + {\tan ^3}\alpha ;\)

c) \(\sqrt {{{\sin }^2}\alpha \left( {1 + \cot \alpha } \right) + {{\cos }^2}\alpha \left( {1 + \tan \alpha } \right)}  = \left| {\sin \alpha  + \cos \alpha } \right|\)

d) \({\sin ^2}\alpha {\tan ^2}\alpha  + 4{\sin ^2}\alpha  - {\tan ^2}\alpha  + 3{\cos ^2}\alpha  = 3\).

(Giả sử các biểu thức đã cho đều có nghĩa).

Giải:

a)

\(\begin{array}{l}\dfrac{{{{\tan }^2}\alpha  - {{\sin }^2}\alpha }}{{{{\cot }^2}\alpha  - {{\cos }^2}\alpha }} = \dfrac{{{{\sin }^2}\alpha \left( {\dfrac{1}{{{{\cos }^2}\alpha }} - 1} \right)}}{{{{\cos }^2}\alpha \left( {\dfrac{1}{{{{\sin }^2}\alpha }} - 1} \right)}}\\ = \dfrac{{{{\sin }^2}\alpha {{\tan }^2}\alpha }}{{{{\cos }^2}\alpha {{\cot }^2}\alpha }} = {\tan ^6}\alpha \end{array}\)

b)

\(\begin{array}{l}\dfrac{{\sin \alpha  + \cos \alpha }}{{{{\cos }^3}\alpha }} = \dfrac{{\cos \alpha \left( {\tan \alpha  + 1} \right)}}{{{{\cos }^3}\alpha }}\\ = \left( {{{\tan }^2}\alpha  + 1} \right)\left( {\tan \alpha  + 1} \right)\\ = 1 + \tan \alpha  + {\tan ^2}\alpha  + {\tan ^3}\alpha .\end{array}\)

c)

\(\begin{array}{l}\sqrt {{{\sin }^2}\alpha \left( {1 + \cot \alpha } \right) + {{\cos }^2}\alpha \left( {1 + \tan \alpha } \right)} \\ = \sqrt {{{\sin }^2}\alpha  + \sin \alpha \cos \alpha  + {{\cos }^2}\alpha  + \cos \alpha \sin \alpha } \\ = \sqrt {{{\left( {\sin \alpha  + \cos \alpha } \right)}^2}}  = \left| {\sin \alpha  + \cos \alpha } \right|.\end{array}\)

d) \(\begin{array}{l}{\sin ^2}\alpha {\tan ^2}\alpha  + 4{\sin ^2}\alpha  - {\tan ^2}\alpha  + 3{\cos ^2}\alpha \\ =  - {\tan ^2}\alpha {\cos ^2}\alpha  + 4{\sin ^2}\alpha  + 3{\cos ^2}\alpha \\ = 3\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) = 3\end{array}\)

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