If `y=tan^(-1) ("cosec x"-cot x),"then " dy/dx=`

Options

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

Answer

To find the derivative of the function \( y = \tan^{-1}(\csc x - \cot x) \), we will follow these steps:

### Step 1: Differentiate the outer function
We start by applying the derivative of the inverse tangent function. The derivative of \( \tan^{-1}(u) \) is given by:

\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

where \( u = \csc x - \cot x \).

### Step 2: Differentiate the inner function
Next, we need to differentiate \( u = \csc x - \cot x \). We recall the derivatives of \( \csc x \) and \( \cot x \):

- The derivative of \( \csc x \) is \( -\csc x \cot x \).
- The derivative of \( \cot x \) is \( -\csc^2 x \).

Thus, we have:

\[ \frac{du}{dx} = \frac{d}{dx}(\csc x) - \frac{d}{dx}(\cot x) = -\csc x \cot x + \csc^2 x \]

### Step 3: Substitute back into the derivative formula
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative formula:

\[ \frac{dy}{dx} = \frac{1}{1 + (\csc x - \cot x)^2} \cdot \left(-\csc x \cot x + \csc^2 x\right) \]

### Step 4: Simplify the expression
We can simplify the expression further. First, we can factor out \( -\csc x \):

\[ \frac{dy}{dx} = \frac{-\csc x (\cot x - \csc x)}{1 + (\csc x - \cot x)^2} \]

Next, we simplify the denominator using the identity \( 1 + \cot^2 x = \csc^2 x \):

\[ 1 + (\csc x - \cot x)^2 = 1 + (\csc^2 x - 2\csc x \cot x + \cot^2 x) = \csc^2 x + 1 - 2\csc x \cot x \]

Thus, we can rewrite our derivative as:

\[ \frac{dy}{dx} = \frac{-\csc x (\cot x - \csc x)}{2\csc^2 x - 2\csc x \cot x} \]

### Final Answer:
The final expression for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{-\csc x (\cot x - \csc x)}{2(\csc^2 x - \csc x \cot x)} \]