Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\tan \alpha + \cot \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\sin \alpha .\cos \alpha }} = \dfrac{1}{{\sin \alpha .\cos \alpha }}\\
b,\\
\dfrac{{\tan \alpha }}{{\cos \alpha }} - \dfrac{{\cos \alpha }}{{\cot \alpha }} = \dfrac{{\tan \alpha .\cot \alpha - co{s^2}\alpha }}{{\cos \alpha .\cot \alpha }}\\
= \dfrac{{1 - {{\cos }^2}\alpha }}{{\cos \alpha .\dfrac{{\cos \alpha }}{{\sin \alpha }}}} = \dfrac{{{{\sin }^2}\alpha }}{{\dfrac{{{{\cos }^2}\alpha }}{{\sin \alpha }}}} = \sin \alpha .\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \sin \alpha .ta{n^2}\alpha \\
c,\\
\sin \left( {a + b} \right).\sin \left( {a - b} \right)\\
= - \dfrac{1}{2}.\left[ {\cos \left( {a + b + a - b} \right) - \cos \left( {a + b - a + b} \right)} \right]\\
= - \dfrac{1}{2}\left[ {\cos 2a - \cos 2b} \right]\\
= - \dfrac{1}{2}.\left[ {\left( {1 - 2{{\sin }^2}a} \right) - \left( {1 - 2{{\sin }^2}b} \right)} \right]\\
= {\sin ^2}a - {\sin ^2}b
\end{array}\)
