Find `("cos"(2pi+theta)"c o s e c"(2pi+theta)"tan"(pi//2+theta))/("sec"(pi//2+theta)costheta"cot"(pi+theta))`

Options

A. 0
B. 1
C. -1
D. 2

Explanation

To solve the expression

\[ \frac{\cos(2\pi + \theta) \cdot \csc(2\pi + \theta) \cdot \tan\left(\frac{\pi}{2} + \theta\right)}{\sec\left(\frac{\pi}{2} + \theta\right) \cdot \cos\theta \cdot \cot(\pi + \theta)} \]

we will simplify both the numerator and the denominator step by step.

### Step 1: Simplify the numerator

1. **Evaluate \(\cos(2\pi + \theta)\)**:
\[ \cos(2\pi + \theta) = \cos(\theta) \quad \text{(since cosine is periodic with period } 2\pi\text{)} \]

2. **Evaluate \(\csc(2\pi + \theta)\)**:
\[ \csc(2\pi + \theta) = \csc(\theta) \quad \text{(since cosecant is also periodic with period } 2\pi\text{)} \]

3. **Evaluate \(\tan\left(\frac{\pi}{2} + \theta\right)\)**:
\[ \tan\left(\frac{\pi}{2} + \theta\right) = -\cot(\theta) \quad \text{(since tangent shifts by } \frac{\pi}{2}\text{)} \]

Putting it all together, the numerator becomes:
\[ \cos(2\pi + \theta) \cdot \csc(2\pi + \theta) \cdot \tan\left(\frac{\pi}{2} + \theta\right) = \cos(\theta) \cdot \csc(\theta) \cdot (-\cot(\theta)) \]

### Step 2: Simplify the denominator

1. **Evaluate \(\sec\left(\frac{\pi}{2} + \theta\right)\)**:
\[ \sec\left(\frac{\pi}{2} + \theta\right) = -\csc(\theta) \quad \text{(since secant shifts by } \frac{\pi}{2}\text{)} \]

2. **Evaluate \(\cot(\pi + \theta)\)**:
\[ \cot(\pi + \theta) = -\cot(\theta) \quad \text{(since cotangent is periodic with period } \pi\text{)} \]

Putting it all together, the denominator becomes:
\[ \sec\left(\frac{\pi}{2} + \theta\right) \cdot \cos\theta \cdot \cot(\pi + \theta) = (-\csc(\theta)) \cdot \cos\theta \cdot (-\cot(\theta)) \]

### Step 3: Combine the results

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

### Step 4: Simplify the fraction

The negative signs cancel out, and we have:
\[ \frac{\cos(\theta) \cdot \csc(\theta) \cdot \cot(\theta)}{\csc(\theta) \cdot \cos\theta \cdot \cot(\theta)} \]

Since the numerator and denominator are identical, we can simplify this to:
\[ 1 \]

### Final Result
Thus, we have shown that:
\[ \frac{\cos(2\pi + \theta) \cdot \csc(2\pi + \theta) \cdot \tan\left(\frac{\pi}{2} + \theta\right)}{\sec\left(\frac{\pi}{2} + \theta\right) \cdot \cos\theta \cdot \cot(\pi + \theta)} = 1 \]