(a) Starting with the identity $\sin^2(x)+\cos^2(x)=1$, divide each side by $\sin^2(x)$. What is the resulting identity?

A. $1+\cot^2(x)=\sec^2(x)$
B. $1+\tan^2(x)=\sec^2(x)$
C. $\csc^2(x)+\cot^2(x)=1$
D. $1+\sec^2(x)=\cot^2(x)$
E. $1+\sec^2(x)=\tan^2(x)$
F. $1+\cot^2(x)=\csc^2(x)$
G. $\cot^2(x)+\sec^2(x)=1$
H. $\sec^2(x)+\tan^2(x)=1$
I. $1+\csc^2(x)=\cot^2(x)$
J. None of the above

(b) Let's prove the identity $\cos^2(x)+(\csc(x)-\sin(x))^2=\cot^2(x)$. What is a good first step to show both sides are equal?

A. $\cos^2(x)+\csc^2(x)+2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)$
B. $\cos^2(x)+\csc^2(x)+\sin^2(x)=\cot^2(x)$
C. $\cos^2(x)+\csc^2(x)-2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)$
D. None of the above

(c) We can use the Pythagorean Identity to simplify the previous equation. What does doing so result in?

A. $-\sin^2(x)+\csc^2(x)+\sin^2(x)=\cot^2(x)$
B. $1+\csc^2(x)+2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)$
C. $\csc^2(x)+2\csc(x)\sin(x)=\cot^2(x)$
D. $1+\csc^2(x)-2\csc(x)\sin(x)=\cot^2(x)$
E. $\cos^2(x)+\csc^2(x)-2\csc(x)\sin(x)+\sin^2(x)=\cot^2(x)$
F. None of the above

(d) Using the definition of $\csc(x)$, we can rewrite the previous equation as...

A. $-\sin^2(x)+\cot^2(x)+\sin^2(x)=\cot^2(x)$
B. $\csc^2(x)+2\csc(x)\sin(x)=\cot^2(x)$
C. $1+\csc^2(x)+2\cot(x)=\cot^2(x)$
D. $1+\csc^2(x)-2=\cot^2(x)$
E. $\csc^2(x)+2\sin^2(x)=\cot^2(x)$
F. $1+\csc^2(x)+2=\cot^2(x)$
G. None of the above

(e) The previous simplifies to...

A. $\cot^2(x)=\cot^2(x)$
B. $3+\csc^2(x)=\cot^2(x)$
C. $\csc^2(x)=\cot^2(x)$
D. $\csc^2(x)-1=\cot^2(x)$
E. $1+\csc^2(x)=-\cot^2(x)$
F. $\csc^2(x)+2=\cot^2(x)$
G. None of the above

(f) How can we finish the proof?

A. This equation is only one off from the Pythagorean Identity, so we can apply that here.
B. The equation is true because both sides are identical.
C. This equation is the negative of the identity we proved in part (a)
D. This equation is true by the definition of the cotangent and cosecant functions.
E. Add one to both sides to obtain the identity we proved in part (a).
F. None of the above