$$ \frac{\cos (180^{\circ }+ heta ). an (180^{\circ }- heta )}{\sin (- heta )} $$

Question

$$ \frac{\cos (180^{\circ }+	heta ).	an (180^{\circ }-	heta )}{\sin (-	heta )} $$

UpStudy AI · Accepted Answer

We begin with the expression

$$
\frac{\cos(180^\circ+\theta) \cdot \tan(180^\circ-\theta)}{\sin(-\theta)}
$$

**Step 1. Simplify each trigonometric function.**

1. For the cosine term:
   $$
   \cos(180^\circ + \theta) = -\cos\theta
   $$

2. For the tangent term:
   $$
   \tan(180^\circ - \theta) = -\tan\theta
   $$

3. For the sine term in the denominator:
   $$
   \sin(-\theta) = -\sin\theta
   $$

**Step 2. Substitute the simplified expressions back into the original expression.**

Substitute to get:

$$
\frac{(-\cos\theta) \cdot (-\tan\theta)}{-\sin\theta}
$$

**Step 3. Simplify the numerator.**

The product in the numerator is:

$$
(-\cos\theta) \cdot (-\tan\theta) = \cos\theta \cdot \tan\theta
$$

So the expression becomes:

$$
\frac{\cos\theta \cdot \tan\theta}{-\sin\theta}
$$

**Step 4. Express $$\tan\theta$$ in terms of $$\sin\theta$$ and $$\cos\theta$$.**

Recall that:

$$
\tan\theta = \frac{\sin\theta}{\cos\theta}
$$

Substitute this into the expression:

$$
\frac{\cos\theta \cdot \frac{\sin\theta}{\cos\theta}}{-\sin\theta}
$$

**Step 5. Cancel common factors.**

The $$\cos\theta$$ terms cancel, leaving:

$$
\frac{\sin\theta}{-\sin\theta}
$$

Assuming $$\sin\theta \neq 0$$, the fraction simplifies to:

$$
-1
$$

Thus, the final simplified expression is:

$$
-1
$$
The expression simplifies to $$-1$$.

\( \frac{\cos (180^{\circ }+\theta ).\tan (180^{\circ }-\theta )}{\sin (-\theta )} \)