If `tan(cotx)=cot(tanx), then sin2x` is equal to

Options

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

Answer

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

### Step 1: Rewrite the given equation
We start with the equation:
\[ \tan(\cot x) = \cot(\tan x) \]

### Step 2: Use the cotangent identity
Recall that:
\[ \cot \theta = \tan\left(\frac{\pi}{2} - \theta\right) \]
Thus, we can rewrite \( \cot(\tan x) \) as:
\[ \cot(\tan x) = \tan\left(\frac{\pi}{2} - \tan x\right) \]
So, we have:
\[ \tan(\cot x) = \tan\left(\frac{\pi}{2} - \tan x\right) \]

### Step 3: Set the angles equal
Since \( \tan A = \tan B \) implies:
\[ A = n\pi + B \quad \text{for some integer } n \]
We can set:
\[ \cot x = n\pi + \left(\frac{\pi}{2} - \tan x\right) \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \cot x + \tan x = n\pi + \frac{\pi}{2} \]

### Step 5: Convert to sine and cosine
Using the identities \( \cot x = \frac{\cos x}{\sin x} \) and \( \tan x = \frac{\sin x}{\cos x} \), we can rewrite the equation:
\[ \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = n\pi + \frac{\pi}{2} \]
Combining the fractions:
\[ \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = n\pi + \frac{\pi}{2} \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \frac{1}{\sin x \cos x} = n\pi + \frac{\pi}{2} \]

### Step 6: Solve for \( \sin 2x \)
Recall that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can express \( \sin 2x \) in terms of the previous equation:
\[ \sin 2x = \frac{2}{n\pi + \frac{\pi}{2}} \]

### Step 7: Final expression for \( \sin 2x \)
To simplify:
\[ \sin 2x = \frac{4}{2n + 1}\pi \]

### Conclusion
Thus, the final answer for \( \sin 2x \) is:
\[ \sin 2x = \frac{4}{(2n + 1)\pi} \]