`cosec^(-1)((1+tan ^2x)/(2tanx))`

Answer

To solve the problem of differentiating the expression \( y = \csc^{-1}\left(\frac{1 + \tan^2 x}{2 \tan x}\right) \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Expression**:
We start with the expression:
\[ y = \csc^{-1}\left(\frac{1 + \tan^2 x}{2 \tan x}\right) \]

2. **Simplify the Argument**:
Recall the trigonometric identity:
\[ 1 + \tan^2 x = \sec^2 x \]
Therefore, we can rewrite the argument of the cosecant inverse:
\[ \frac{1 + \tan^2 x}{2 \tan x} = \frac{\sec^2 x}{2 \tan x} \]

3. **Rewrite in Terms of Sine and Cosine**:
We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can rewrite:
\[ \sec^2 x = \frac{1}{\cos^2 x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Therefore, the expression becomes:
\[ \frac{\frac{1}{\cos^2 x}}{2 \cdot \frac{\sin x}{\cos x}} = \frac{1}{2 \sin x \cos x} \]

4. **Use the Double Angle Identity**:
We know that:
\[ 2 \sin x \cos x = \sin(2x) \]
So, we can express our argument as:
\[ \frac{1}{\sin(2x)} \]
Hence, we can rewrite \( y \) as:
\[ y = \csc^{-1}(\csc(2x)) \]

5. **Simplify the Inverse Cosecant**:
Since \( \csc^{-1}(\csc(2x)) = 2x \) (for \( x \) in the appropriate range), we have:
\[ y = 2x \]

6. **Differentiate with Respect to x**:
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x) = 2 \]

### Final Answer:
The derivative of the given expression is:
\[ \frac{dy}{dx} = 2 \]