Đáp án:
\({\left[ {\begin{array}{*{20}{l}}
{A = \dfrac{{10}}{{\sqrt 7 }}}\\
{A = - \dfrac{{10}}{{\sqrt 7 }}{\rm{\;}}}
\end{array}} \right.}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{A = \dfrac{{{{\sin }^2}a - {{\cos }^2}a}}{{2\sin a\cos a}} - \cot a}\\
{ = \dfrac{{1 - {{\cos }^2}a - {{\cos }^2}a}}{{2\sin a\cos a}} - \dfrac{{\cos a}}{{\sin a}}}\\
{ = \dfrac{{1 - 2{{\cos }^2}a - 2{{\cos }^2}a}}{{2\sin a\cos a}}}\\
{ = \dfrac{{1 - 4{{\cos }^2}a}}{{2\sin a\cos a}}}\\
{Do:\cos a = \dfrac{1}{8}}\\
{{{\cos }^2}a + {{\sin }^2}a = 1}\\
{ \to \dfrac{1}{{64}} + {{\sin }^2}a = 1}\\
{ \to {{\sin }^2}a = \dfrac{{63}}{{64}}}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\sin a = \dfrac{{\sqrt {63} }}{8}}\\
{\sin a = - \dfrac{{\sqrt {63} }}{8}}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\dfrac{{1 - 4{{\cos }^2}a}}{{2\sin a\cos a}} = \dfrac{{1 - 4.\dfrac{1}{{64}}}}{{2.\dfrac{{\sqrt {63} }}{8}.\dfrac{1}{8}}}}\\
{\dfrac{{1 - 4{{\cos }^2}a}}{{2\sin a\cos a}} = \dfrac{{1 - 4.\dfrac{1}{{64}}}}{{2.\left( { - \dfrac{{\sqrt {63} }}{8}} \right).\dfrac{1}{8}}}{\rm{\;}}}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{A = \dfrac{{10}}{{\sqrt 7 }}}\\
{A = - \dfrac{{10}}{{\sqrt 7 }}{\rm{\;}}}
\end{array}} \right.}
\end{array}\)
