Find the value of `(sin(180^0+theta)cos(90^0+theta)t a n(270^0-theta))/(sin(180^0-theta)"sin"(270^0+theta))`

Options

A. 0
B. 1
C. -1
D. None of these

Explanation

To find the value of

\[ \frac{\sin(180^\circ + \theta) \cos(90^\circ + \theta) \tan(270^\circ - \theta)}{\sin(180^\circ - \theta) \sin(270^\circ + \theta)} \]

we will use trigonometric identities and properties of angles.

### Step 1: Simplify the numerator

1. **Calculate \(\sin(180^\circ + \theta)\)**:
\[ \sin(180^\circ + \theta) = -\sin(\theta) \]

2. **Calculate \(\cos(90^\circ + \theta)\)**:
\[ \cos(90^\circ + \theta) = -\sin(\theta) \]

3. **Calculate \(\tan(270^\circ - \theta)\)**:
\[ \tan(270^\circ - \theta) = -\cot(\theta) \]

Now, substituting these values into the numerator:
\[ \text{Numerator} = (-\sin(\theta)) \cdot (-\sin(\theta)) \cdot (-\cot(\theta)) = -\sin^2(\theta) \cot(\theta) \]

### Step 2: Simplify the denominator

1. **Calculate \(\sin(180^\circ - \theta)\)**:
\[ \sin(180^\circ - \theta) = \sin(\theta) \]

2. **Calculate \(\sin(270^\circ + \theta)\)**:
\[ \sin(270^\circ + \theta) = -\cos(\theta) \]

Now, substituting these values into the denominator:
\[ \text{Denominator} = \sin(\theta) \cdot (-\cos(\theta)) = -\sin(\theta) \cos(\theta) \]

### Step 3: Combine the results

Now we can substitute the simplified numerator and denominator back into the original expression:
\[ \frac{-\sin^2(\theta) \cot(\theta)}{-\sin(\theta) \cos(\theta)} = \frac{\sin^2(\theta) \cot(\theta)}{\sin(\theta) \cos(\theta)} \]

### Step 4: Further simplification

Since \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\), we can write:
\[ \frac{\sin^2(\theta) \cdot \frac{\cos(\theta)}{\sin(\theta)}}{\sin(\theta) \cos(\theta)} = \frac{\sin(\theta) \cos(\theta)}{\sin(\theta) \cos(\theta)} = 1 \]

### Conclusion

Thus, the value of the original expression is:
\[ 1 \]