If ` y = tan^(-1) (sec x - tan x ) , "then" (dy)/(dx)` is equal to

Answer

To solve the problem, we need to find the derivative \( \frac{dy}{dx} \) for the function \( y = \tan^{-1}(\sec x - \tan x) \).

### Step-by-Step Solution:

1. **Rewrite the Function**:
We start with the given function:
\[ y = \tan^{-1}(\sec x - \tan x) \]

2. **Use Trigonometric Identities**:
Recall the identities:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Therefore, we can rewrite \( \sec x - \tan x \) as:
\[ \sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x} \]

3. **Substitute Back into the Function**:
Now, substituting this back into our function for \( y \):
\[ y = \tan^{-1}\left(\frac{1 - \sin x}{\cos x}\right) \]

4. **Differentiate Using the Chain Rule**:
To find \( \frac{dy}{dx} \), we apply the chain rule:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 - \sin x}{\cos x}\right)^2} \cdot \frac{d}{dx}\left(\frac{1 - \sin x}{\cos x}\right) \]

5. **Differentiate the Inner Function**:
We need to differentiate \( \frac{1 - \sin x}{\cos x} \) using the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 - \sin x}{\cos x}\right) = \frac{\cos x \cdot (-\cos x) - (1 - \sin x)(-\sin x)}{\cos^2 x} \]
Simplifying this gives:
\[ = \frac{-\cos^2 x + (1 - \sin x)\sin x}{\cos^2 x} \]

6. **Combine the Results**:
Now, substitute this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 - \sin x}{\cos x}\right)^2} \cdot \frac{-\cos^2 x + (1 - \sin x)\sin x}{\cos^2 x} \]

7. **Simplify**:
After simplification, we can evaluate the expression to find \( \frac{dy}{dx} \).

8. **Final Result**:
After performing the calculations, we find:
\[ \frac{dy}{dx} = -\frac{1}{2} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{1}{2} \]