Consider the function $f(x) = \tan(x)$, and remember that $\tan(x) = \displaystyle{\frac{\sin(x)}{\cos(x)}}$.

(a) What is the domain of $f$?
A. All real numbers except $\pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, \ldots$
R. All real numbers
B. All real numbers except zero
C. All real numbers except $0, \pm\pi, \pm 2\pi, \pm 3\pi, \ldots$
D. All positive real numbers.
E. None of these

(b) Andre computes $f'(x)$ using the quotient rule. Which of the following is a possible correct answer?
A. $\displaystyle{f'(x)=\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos^2(x)}}$
B. $\displaystyle{f'(x)=\frac{\cos(x)\cos(x)-\sin(x)\sin(x)}{\cos^2(x)}}$
C. $\displaystyle{f'(x)=\frac{\cos(x)\sin(x)+\sin(x)\cos(x)}{\cos^2(x)}}$
D. $\displaystyle{f'(x)=\frac{\cos(x)\sin(x)-\sin(x)\cos(x)}{\cos^2(x)}}$
E. $\displaystyle{f'(x)=\frac{\sin(x)\sin(x)-\cos(x)\cos(x)}{\cos^2(x)}}$
F. None of these

(c) Berenice simplifies Andre's correct answer using the Fundamental Trigonometric Identity. Which of the following is a correct answer?
A. $\displaystyle{f'(x)=\frac{1}{\cos^2(x)}}$
B. $\displaystyle{f'(x)=\frac{\cos(2x)\sin(2x)}{\cos^2(x)}}$
C. $\displaystyle{f'(x)=\frac{2\cos(x)\sin(x)}{\cos^2(x)}}$
D. $\displaystyle{f'(x)=\frac{\cos(x)-\sin(x)}{\cos^2(x)}}$
E. $\displaystyle{f'(x)=\frac{\sin(x)-\cos(x)}{\cos^2(x)}}$
F. None of these

(d) Corey simplifies Berenice's correct answer using another trigonometric Identity. Which of the following is a correct answer?
A. $f'(x) = \sec^2(x)$
B. $f'(x) = \csc^2(x)$
C. $f'(x) = \cot(x)$
D. $f'(x) = \cot^2(x)$
E. $f'(x) = \csc(x)$
F. $f'(x) = \sec(x)$
G. None of these

(e) For what values of $x$ is $f'(x)$ defined?
A. All real numbers except $\pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, \ldots$
R. All real numbers
B. All real numbers except zero
C. All real numbers except $0, \pm\pi, \pm 2\pi, \pm 3\pi, \ldots$
D. All positive real numbers.
E. None of these