Let's verify the identity $\frac{\sec(\theta)+\csc(\theta)}{1+\tan(\theta)}=\csc(\theta)$. What is a good first step?

A. $\sec(\theta)+\csc(\theta)=\tan(\theta)\csc(\theta)+\tan(\theta)\csc(\theta)$
B. $\frac{\sec(\theta)+\csc(\theta)}{\sec^2(\theta)}=\csc(\theta)$
C. $\sec(\theta)+\csc(\theta)=\csc(\theta)+\tan(\theta)\csc(\theta)$
D. $\frac{\sec(\theta)+\csc(\theta)}{\sec(\theta)}=\csc(\theta)$
E. $\sec(\theta)+\csc(\theta)=1+\tan(\theta)\csc(\theta)$
F. None of the above

What is a quick way to simplify the previous equation?

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

What do we get when we rewrite the right-hand side purely in terms of sine and cosine?

A. $\sec(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$
B. $\cot(\theta)=2\frac{\sin(\theta)}{\cos(\theta)}$
C. $\sec(\theta)+\csc(\theta)=\frac{\sin(\theta)}{\cos^2(\theta)}$
D. $\sec(\theta)=\frac{\sin(\theta)}{\cos(\theta)\sin(\theta)}$
E. $\sec(\theta)=\frac{\sin(\theta)}{\cos^2(\theta)}$
F. $\sec(\theta)+\csc(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$
G. $\cot(\theta)=1+\frac{\sin(\theta)}{\cos(\theta)}$
H. None of the above

How can we finish the proof?

A. We apply the Pythagorean Identity to make both sides equal.
B. We apply the double angle formula for cosine after simplifying.
C. After simplifying the previous equation, we are left with the definition of secant.
D. After simplifying, we get the get double angle formula for sine.
E. None of the above