`tan^(-1)((cosx+sinx)/(cos x-sinx))`

Answer

To find the derivative of the function \( y = \tan^{-1}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) \), we can follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent
We start with the expression inside the inverse tangent:
\[ y = \tan^{-1}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) \]

### Step 2: Rewrite the Expression
We can rewrite the fraction:
\[ \frac{\cos x + \sin x}{\cos x - \sin x} = \frac{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}} = \frac{1 + \tan x}{1 - \tan x} \]
Thus, we have:
\[ y = \tan^{-1}\left(\frac{1 + \tan x}{1 - \tan x}\right) \]

### Step 3: Use the Angle Addition Formula
Using the tangent addition formula:
\[ \tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4} \tan x} = \frac{1 + \tan x}{1 - \tan x} \]
This means:
\[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + x\right)\right) \]

### Step 4: Simplify Using the Inverse Function
Since the tangent and inverse tangent functions are inverses, we can simplify:
\[ y = \frac{\pi}{4} + x \]

### Step 5: Differentiate
Now, we differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = 0 + 1 = 1 \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = 1 \]

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