Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
\dfrac{{\tan \alpha - \sin \alpha }}{{{{\sin }^3}\alpha }} = \dfrac{{\dfrac{{\sin \alpha }}{{\cos \alpha }} - \sin \alpha }}{{{{\sin }^3}\alpha }} = \dfrac{{\sin \alpha .\left( {\dfrac{1}{{\cos \alpha }} - 1} \right)}}{{{{\sin }^3}\alpha }}\\
= \dfrac{{\dfrac{{1 - \cos \alpha }}{{\cos \alpha }}}}{{{{\sin }^2}\alpha }} = \dfrac{{1 - \cos \alpha }}{{\cos \alpha .\left( {1 - {{\cos }^2}\alpha } \right)}}\\
= \dfrac{{1 - \cos \alpha }}{{\cos \alpha \left( {1 - \cos \alpha } \right)\left( {1 + \cos \alpha } \right)}} = \dfrac{1}{{\cos \alpha \left( {1 + \cos \alpha } \right)}}\\
f,\\
\dfrac{{{{\tan }^2}2\alpha - {{\tan }^2}\alpha }}{{1 - {{\tan }^2}2\alpha .ta{n^2}\alpha }}\\
= \dfrac{{\left( {\tan 2\alpha - \tan \alpha } \right)\left( {\tan 2\alpha + \tan \alpha } \right)}}{{\left( {1 - \tan 2\alpha .\tan \alpha } \right)\left( {1 + \tan 2\alpha .\tan \alpha } \right)}}\\
= \dfrac{{\tan 2\alpha - \tan \alpha }}{{1 + \tan 2\alpha .\tan \alpha }}.\dfrac{{\tan 2\alpha + \tan \alpha }}{{1 - \tan 2\alpha .\tan \alpha }}\\
= \tan \left( {2\alpha - \alpha } \right).\tan \left( {2\alpha + \alpha } \right)\\
= \tan \alpha .\tan 3\alpha
\end{array}\)
Em xem lại đề câu E nhé, sai rồi!
