Multiple Choice 1 point Considering only the values of $$ \alpha $$ for which $$ an \alpha \cos (2 \alpha) $$ is defined, which of the following expressions is equivalent to tan $$ \alpha \cos (2 \alpha) $$ ? $$ 2 \cos ^{2} \alpha $$ $$ \cos \alpha \cot \alpha-\sin \alpha $$ $$ \frac{2 \sec \alpha}{1- an ^{2} \alpha} $$ $$ 2 \sin \alpha \cos \alpha- an \alpha $$ $$ 2 \sin ^{2} \alpha- an ^{2} \alpha $$

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Multiple Choice 1 point Considering only the values of $$ \alpha $$ for which $$ 	an \alpha \cos (2 \alpha) $$ is defined, which of the following expressions is equivalent to tan $$ \alpha \cos (2 \alpha) $$ ? $$ 2 \cos ^{2} \alpha $$ $$ \cos \alpha \cot \alpha-\sin \alpha $$ $$ \frac{2 \sec \alpha}{1-	an ^{2} \alpha} $$ $$ 2 \sin \alpha \cos \alpha-	an \alpha $$ $$ 2 \sin ^{2} \alpha-	an ^{2} \alpha $$

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We start with the expression tan α cos 2α and recall that tan α = sin α/cos α. Thus we have

tan α cos 2α = (sin α/cos α) cos 2α = (sin α cos 2α)/cos α.

Now, consider option 4: 2 sin α cos α - tan α. We rewrite tan α in this expression as sin α/cos α:

2 sin α cos α - tan α = 2 sin α cos α - (sin α/cos α).

To combine these terms with a common denominator, express 2 sin α cos α as (2 sin α cos² α)/cos α:

= (2 sin α cos² α - sin α)/cos α
  = sin α (2 cos² α - 1)/cos α.

Recall that the double-angle identity for cosine is cos 2α = 2 cos² α - 1. Therefore,

sin α (2 cos² α - 1)/cos α = (sin α cos 2α)/cos α.

This is exactly the expression we derived for tan α cos 2α. Hence, option 4 is equivalent to tan α cos 2α.

Thus, the correct answer is:

2 sin α cos α - tan α.
The correct expression equivalent to tan α cos 2α is $$2 \sin \alpha \cos \alpha - 	an \alpha$$.

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