Answer
$$\int^{0}_{-\pi/3}\sec x\tan xdx=-1$$
Work Step by Step
$$A=\int^{0}_{-\pi/3}\sec x\tan xdx$$ $$A=\sec x\Big]^{0}_{-\pi/3}=\frac{1}{\cos x}\Big]^{0}_{-\pi/3}$$ $$A=\frac{1}{\cos0}-\frac{1}{\cos(-\pi/3)}$$ $$A=1-\frac{1}{\frac{1}{2}}$$ $$A=1-2=-1$$
