Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( {\dfrac{1}{4}} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{{15}}{{16}}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos a = \dfrac{{\sqrt {15} }}{4} \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{1}{{\sqrt {15} }};\,\,\,\cot a = \dfrac{1}{{\tan a}} = \sqrt {15} \\
\cos a = - \dfrac{{\sqrt {15} }}{4} \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{{ - 1}}{{\sqrt {15} }};\,\,\,\cot a = \dfrac{1}{{\tan a}} = - \sqrt {15}
\end{array} \right.\\
b,\\
\sin x = \cos \left( {90^\circ - x} \right)\\
A = \dfrac{{\sin 43^\circ }}{{\cos 47^\circ }} + \tan 45^\circ - \cos 20^\circ .\cot 70^\circ \\
= \dfrac{{\cos \left( {90^\circ - 43^\circ } \right)}}{{\cos 47^\circ }} + \dfrac{{\sin 45^\circ }}{{\cos 45^\circ }} - \cos 20^\circ .\dfrac{{\cos 70^\circ }}{{\sin 70^\circ }}\\
= \dfrac{{\cos 47^\circ }}{{\cos 47^\circ }} + \dfrac{{\cos \left( {90^\circ - 45^\circ } \right)}}{{\cos 45^\circ }} - \cos 20^\circ .\dfrac{{\cos 70^\circ }}{{\cos \left( {90^\circ - 70^\circ } \right)}}\\
= 1 + \dfrac{{\cos 45^\circ }}{{\cos 45^\circ }} - \cos 20^\circ .\dfrac{{\cos 70^\circ }}{{\cos 20^\circ }}\\
= 1 + 1 - \cos 70^\circ \\
= 2 - \cos 70^\circ
\end{array}\)
