\(A= \dfrac{8{\tan}^2 a + 3\cot a -1}{\tan a + \cot a}\)
\(=\dfrac{\dfrac{{8\sin }^2a}{{\cos}^2a}+\dfrac{3\cos a}{\sin a}-1}{\dfrac{\sin a}{\cos a}+\dfrac{\cos a}{\sin a}}\)
Ta có: \({\sin }^2 x+{\cos}^2x=1\)
Với \(\sin a=\dfrac{1}{3}\) do \(90^o
\(\Rightarrow \dfrac{\sin a}{\cos a}=-\dfrac{1}{2\sqrt 2}\) và \(\dfrac{\cos a}{\sin a}=-2\sqrt 2\)
\(\Rightarrow A=\dfrac{\dfrac{{8\sin }^2a}{{\cos}^2a}+\dfrac{3\cos a}{\sin a}-1}{\dfrac{\sin a}{\cos a}+\dfrac{\cos a}{\sin a}}=\dfrac{8.\dfrac{1}{8}-3.2\sqrt 2-1}{-\dfrac{1}{2\sqrt 2}-2\sqrt 2}=\dfrac{6\sqrt 2}{\dfrac{1+8}{2\sqrt 2}}=\dfrac{6\sqrt 2.2\sqrt 2}{9}=\dfrac{8}{3}\)
