Question

An incomplete proof of 1 minus start fraction tangent theta cosine theta over cosecant theta end fraction equals cosine squared theta is shown.

math expression1 minus start fraction tangent theta cosine theta over cosecant theta end fraction equals 1 minus start fraction start fraction sine theta over cosine theta end fraction cosine theta over cosecant theta end fraction line break
1 minus start fraction tangent theta cosine theta over cosecant theta end fraction equals line break
1 minus start fraction tangent theta cosine theta over cosecant theta end fraction equals 1 minus sine theta sine theta line break
1 minus start fraction tangent theta cosine theta over cosecant theta end fraction equals 1 minus sine squared theta line break
1 minus start fraction tangent theta cosine theta over cosecant theta end fraction equals cosine squared theta

Which of the following expressions could be used to complete the second step of the proof?
(1 point)
Responses

1 minus cosine theta cosine theta
Image with alt text: 1 minus cosine theta cosine theta

1 minus sine squared theta cosine squared theta
Image with alt text: 1 minus sine squared theta cosine squared theta

1 minus start fraction sine theta cosecant theta over cosine theta end fraction
Image with alt text: 1 minus start fraction sine theta cosecant theta over cosine theta end fraction

1 minus start fraction sine theta over cosecant theta end fraction

Answer 1

To complete the second step of the proof, we want to rewrite the expression \(1 - \frac{\tan \theta \cos \theta}{\csc \theta}\).

Given that:
- \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
- \(\csc \theta = \frac{1}{\sin \theta}\)

We can substitute these definitions into the expression:

\[
1 - \frac{\tan \theta \cos \theta}{\csc \theta} = 1 - \frac{\frac{\sin \theta}{\cos \theta} \cdot \cos \theta}{\frac{1}{\sin \theta}}
\]

This simplifies to:

\[
1 - \frac{\sin \theta}{\cos \theta} \cdot \cos \theta \cdot \sin \theta = 1 - \sin^2 \theta
\]

Since we want to find a suitable expression to replace \(\frac{\tan \theta \cos \theta}{\csc \theta}\) in the proof, let's simplify that step a bit more.

Given this, we need to rewrite \(\frac{\tan \theta \cos \theta}{\csc \theta}\) directly:

\[
\frac{\tan \theta \cos \theta}{\csc \theta} = \tan \theta \cos \theta \cdot \sin \theta = \frac{\sin \theta}{\cos \theta} \cdot \cos \theta \cdot \sin \theta = \sin^2 \theta
\]

Thus, the expression \(1 - \sin^2 \theta\) gives us \(\cos^2 \theta\).

Now examining the provided multiple-choice options, we want one that represents the necessary form:

1. \(1 - \cos^2 \theta\) → not relevant here.
2. \(1 - \sin^2 \theta \cos^2 \theta\) → not matching our requirement.
3. \(1 - \frac{\sin \theta \csc \theta}{\cos \theta}\) → incorrect fraction.
4. \(1 - \frac{\sin \theta}{\csc \theta}\) → simplifies to \(1 - \sin^2 \theta\).

The correct choice that leads to the necessary transformation is:

**\(1 - \frac{\sin \theta}{\csc \theta}\)**.

This choice simplifies directly to \(1 - \sin^2 \theta\), which matches the needed step to reach \(1 - \frac{\tan \theta \cos \theta}{\csc \theta}\).