If `"sin" (pi "cot" theta) = "cos" (pi "tan" theta), "then cosec" 2 theta` is equal to

Options

A. `n - (1)/(2)`
B. `n + (1)/(4)`
C. 4n+1
D. 4n-1

Answer

To solve the equation \( \sin(\pi \cot \theta) = \cos(\pi \tan \theta) \) and find the value of \( \csc(2\theta) \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \sin x = \cos\left(\frac{\pi}{2} - x\right) \]
Thus, we can rewrite the left side of the equation:
\[ \sin(\pi \cot \theta) = \cos\left(\frac{\pi}{2} - \pi \cot \theta\right) \]
So, we have:
\[ \cos\left(\frac{\pi}{2} - \pi \cot \theta\right) = \cos(\pi \tan \theta) \]

### Step 2: Set the arguments equal to each other
Since \( \cos A = \cos B \), we can set the arguments equal to each other:
\[ \frac{\pi}{2} - \pi \cot \theta = \pi \tan \theta + 2n\pi \quad \text{or} \quad \frac{\pi}{2} - \pi \cot \theta = -(\pi \tan \theta) + 2n\pi \]
for some integer \( n \).

### Step 3: Solve the first equation
Taking the first case:
\[ \frac{\pi}{2} - \pi \cot \theta = \pi \tan \theta + 2n\pi \]
Rearranging gives:
\[ \frac{\pi}{2} = \pi \tan \theta + \pi \cot \theta + 2n\pi \]
Dividing through by \( \pi \):
\[ \frac{1}{2} = \tan \theta + \cot \theta + 2n \]
Using the identity \( \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \):
\[ \frac{1}{2} = \frac{1}{\sin \theta \cos \theta} + 2n \]

### Step 4: Rearranging to find \( \sin \theta \cos \theta \)
Rearranging gives:
\[ \sin \theta \cos \theta = \frac{1}{2(2n + 1)} \]
Using the double angle identity \( \sin(2\theta) = 2\sin \theta \cos \theta \):
\[ \sin(2\theta) = \frac{1}{2n + 1} \]

### Step 5: Find \( \csc(2\theta) \)
The cosecant function is the reciprocal of the sine function:
\[ \csc(2\theta) = \frac{1}{\sin(2\theta)} = 2n + 1 \]

### Conclusion
Thus, the value of \( \csc(2\theta) \) is:
\[ \csc(2\theta) = 2n + 1 \]