Establish the identity. $$ \cos \theta(\cot \theta+\tan \theta)=\boldsymbol{\operatorname { c s c }} \theta $$ Write the left side in terms of sine and cosine. $$ \cos \theta(\square) $$

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To establish the identity

$$
\cos \theta(\cot \theta+\tan \theta)=\operatorname{csc} \theta,
$$

we will start by rewriting the left side in terms of sine and cosine.

1. **Rewrite $$\cot \theta$$ and $$\tan \theta$$**:
   - $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$
   - $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

2. **Substituting these into the left side**:
   $$
   \cos \theta \left( \cot \theta + \tan \theta \right) = \cos \theta \left( \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right)
   $$

3. **Combine the terms inside the parentheses**:
   To combine the fractions, we need a common denominator, which is $$\sin \theta \cos \theta$$:
   $$
   \frac{\cos^2 \theta}{\sin \theta} + \frac{\sin^2 \theta}{\cos \theta} = \frac{\cos^2 \theta \cdot \cos \theta + \sin^2 \theta \cdot \sin \theta}{\sin \theta \cos \theta} = \frac{\cos^3 \theta + \sin^2 \theta \cos \theta}{\sin \theta \cos \theta}
   $$

4. **Now multiply by $$\cos \theta$$**:
   $$
   \cos \theta \left( \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right) = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta}
   $$

5. **Using the Pythagorean identity**:
   We know that $$\cos^2 \theta + \sin^2 \theta = 1$$:
   $$
   \frac{1}{\sin \theta} = \operatorname{csc} \theta
   $$

Thus, we have shown that:

$$
\cos \theta(\cot \theta+\tan \theta) = \operatorname{csc} \theta.
$$

The left side in terms of sine and cosine is:

$$
\cos \theta \left( \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right).
$$
The left side of the identity in terms of sine and cosine is: $$ \cos \theta \left( \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right). $$

Establish the identity.

Write the left side in terms of sine and cosine.

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Establish the identity. Write the left side in terms of sine and cosine.