Đáp án:
$B=\dfrac{13\sqrt{2}}{2}+\dfrac{2\sqrt{2}}{9}$
Giải thích các bước giải:
$1+\cot^2 a=\dfrac{1}{\sin^2 a}\\
\Rightarrow \cot^2 a=\dfrac{1}{\sin^2 a}-1\\
=\dfrac{1}{\left (\dfrac{1}{3} \right )^2}-1=8\\
\Rightarrow \cot a=2\sqrt{2}\\
+) \tan a=\dfrac{1}{\cot a}=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{4}\\
\Rightarrow B=3\cot a+2\tan a+\dfrac{1}{\cot a+\tan a}\\
=3.2\sqrt{2}+2.\dfrac{\sqrt{2}}{4}+\dfrac{1}{2\sqrt{2}+\dfrac{\sqrt{2}}{4}}\\
=6\sqrt{2}+\dfrac{\sqrt{2}}{2}+\dfrac{2\sqrt{2}}{9}\\
=\dfrac{13\sqrt{2}}{2}+\dfrac{2\sqrt{2}}{9}$
