c/
$\begin{array}{l} \cos \left( {\dfrac{\pi }{6} + x} \right) + \cos \left( {\dfrac{\pi }{6} - x} \right)\\ = 2\cos \left( {\dfrac{{\dfrac{\pi }{6} + x + \dfrac{\pi }{6} - x}}{2}} \right)\cos \left( {\dfrac{{\dfrac{\pi }{6} + x - \dfrac{\pi }{6} + x}}{2}} \right)\\ = 2\cos \dfrac{\pi }{6}\cos x\\ = 2.\dfrac{{\sqrt 3 }}{2}\cos x = \sqrt 3 \cos x \end{array}$
d/
$\begin{array}{l}
\dfrac{{\cos 2x}}{{1 - \sin 2x}}\\
= \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x}}\\
= \dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\left( {\cos x - \sin x} \right)}^2}}}\\
= \dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}\\
= \dfrac{{\cos x\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)}}{{\cos x\left( {1 - \dfrac{{\sin x}}{{\cos x}}} \right)}}\\
= \dfrac{{1 + \tan x}}{{1 - \tan x}} = \dfrac{{\tan \dfrac{\pi }{4} + \tan x}}{{1 - \tan \dfrac{\pi }{4}\tan x}} = \tan \left( {\dfrac{\pi }{4} + x} \right)\\
\to Sai\,đề
\end{array}$
