Let's verify the identity $\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$. What do we get if we rewrite the right-hand side using sine and cosine?

A. $\tan(2\theta)=\frac{2\frac{\sin(\theta)}{\cos(\theta)}}{1-\frac{\sin^2(\theta)}{\cos^2(\theta)}}$
B. $\tan(2\theta)=\frac{2\frac{\cos(\theta)}{\sin(\theta)}}{1-\frac{\cos^2(\theta)}{\sin^2(\theta)}}$
C. $\tan(2\theta)=\frac{2\frac{\sin(\theta)}{\cos(\theta)}}{1-\frac{\sin(\theta)}{\cos(\theta)}}$
D. $\tan(2\theta)=\frac{2\sin(\theta)\cos(\theta)}{1-\sin^2(\theta)\cos^2(\theta)}$
E. None of the above

What happens when we multiply the right hand side by $\cos^2(\theta)/\cos^2(\theta)$?

A. $\tan(2\theta)=\frac{2\cos(\theta)}{\cos^2(\theta)-\cos^2(\theta)}$
B. $\tan(2\theta)=\frac{\frac{2\sin(\theta)}{\cos(\theta)}}{\cos^2(\theta)-\sin^2(\theta)}$
C. $\tan(2\theta)=\frac{2\sin(\theta)\cos(\theta)}{1-\frac{\sin(\theta)}{cos(\theta)}}$
D. $\tan(2\theta)=\frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)}$
E. None of the above

How can we use double angle formulas to simplify the right hand side?

A. $\tan(2\theta)=\frac{\sin(2\theta)}{\cos(2\theta)}$
B. $\tan(2\theta)=\frac{\cos(2\theta)}{\sin(2\theta)}$
C. $\tan(2\theta)=\sin(2\theta)-\cos(2\theta)$
D. $\tan(2\theta)=\frac{\tan(2\theta)}{\sin^2(\theta)+\cos(2\theta)}$
E. None of the above

How can we finish the proof?

A. The previous equation is true after multiplying both sides by tangent.
B. The previous equation is equivalent to the Pythagorean Theorem.
C. The previous equation is true by the definition of sine and cosine.
D. The previous equation is true by the definition of the double angle formulas.
E. The previous equation is true by the definition of tangent.
F. None of the above