Answer
$A = 2-2\ln 2$
Work Step by Step
From the graph, we can see that on the interval $\big[-\frac{\pi}{3}, 0 \big]$,
$2\sin(x) \leq \tan(x) dx$
and on the interval $\big[0, \frac{\pi}{3} \big]$,
$2\sin(x) \geq \tan(x) dx$;
therefore, the area between the two curves is
$$A = \displaystyle\int_{-\frac{\pi}{3}}^{0} \tan(x) - 2\sin(x)dx + \displaystyle\int_{0}^{\frac{\pi}{3}} 2\sin(x) - \tan(x)dx$$ $$A=\Big[\ln |\cos (x)| -2\cos (x)\Big]^{0}_{-\frac{\pi}{3}} + \Big[-2\cos (x) + \ln |\cos (x)|\Big]^{\frac{\pi}{3}}_0$$ $$A=\big[ -2(1)+\ln(\frac{1}{2})-2(-\frac{1}{2})\big] - \big[-2(\frac{1}{2})+\ln(\frac{1}{2})+2(1)\big]$$ $$A = 2 + \ln(\frac{1}{4})$$ $$A = 2-2\ln 2$$