Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
\tan 15 = \frac{1}{{\cot 15}} = 2 - \sqrt 3 \\
\to \cot 15 = \frac{1}{{2 - \sqrt 3 }} = 2 + \sqrt 3 \\
\tan 15 = \frac{{\sin 15}}{{\cos 15}} = 2 - \sqrt 3 \\
\to \sin 15 = \left( {2 - \sqrt 3 } \right)\cos 15\\
{\sin ^2}15 + {\cos ^2}15 = 1\\
\to {\left( {2 - \sqrt 3 } \right)^2}{\cos ^2}15 + {\cos ^2}15 = 1\\
\to {\cos ^2}15 = \frac{{2 + \sqrt 3 }}{4}\\
\to \cos 15 = \pm \sqrt {\frac{{2 + \sqrt 3 }}{4}} \\
\to \sin 15 = \left( {2 - \sqrt 3 } \right)\cos 15 = \pm \left( {2 - \sqrt 3 } \right).\sqrt {\frac{{2 + \sqrt 3 }}{4}} \\
B2:\\
DK:\tan x \ne \cot x \to \frac{{\sin x}}{{\cos x}} \ne \frac{{\cos x}}{{\sin x}}\\
\to {\cos ^2}x - {\sin ^2}x \ne 0 \to \cos 2x \ne 0\\
\to 2x \ne \frac{\pi }{2} + k\pi \\
\to x \ne \frac{\pi }{4} + \frac{{k\pi }}{2}\\
\sin x = \frac{1}{3}\\
{\sin ^2}x + {\cos ^2}x = 1\\
\to \frac{1}{9} + {\cos ^2}x = 1\\
\to {\cos ^2}x = \frac{8}{9}\\
\to \cos x = \pm \frac{{2\sqrt 2 }}{3}\\
\to \left[ \begin{array}{l}
\tan x = \frac{{\sqrt 2 }}{4}\\
\tan x = - \frac{{\sqrt 2 }}{4}
\end{array} \right. \to \left[ \begin{array}{l}
\cot x = 2\sqrt 2 \\
\cot x = - 2\sqrt 2
\end{array} \right.\\
\to A = - \frac{9}{7}
\end{array}\)
