If `y = tan^(-1){(cos x + sinx)/(cos x - sin x)} " then " (dy)/(dx) =` ?

Options

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

Explanation

To solve the problem, we need to find the derivative of the function given by:

\[ y = \tan^{-1}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) \]

### Step 1: Simplifying the Expression
We can simplify the expression inside the arctangent function. We will factor out \(\cos x\) from both the numerator and the denominator:

\[ y = \tan^{-1}\left(\frac{\cos x(1 + \tan x)}{\cos x(1 - \tan x)}\right) \]

This simplifies to:

\[ y = \tan^{-1}\left(\frac{1 + \tan x}{1 - \tan x}\right) \]

### Step 2: Recognizing the Formula
We can recognize that the expression \(\frac{1 + \tan x}{1 - \tan x}\) can be transformed using the tangent addition formula:

\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]

Here, we can let \(a = \frac{\pi}{4}\) (since \(\tan \frac{\pi}{4} = 1\)) and \(b = x\):

\[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + x\right)\right) \]

Thus, we can simplify \(y\) to:

\[ y = \frac{\pi}{4} + x \]

### Step 3: Differentiating
Now we differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + x\right) \]

The derivative of a constant is zero, and the derivative of \(x\) is 1, so:

\[ \frac{dy}{dx} = 0 + 1 = 1 \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = 1 \]

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