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

Options

A. 0
B. 1
C. `-1`
D. None

Answer

To solve the problem, we need to differentiate the given function \( y = \tan^{-1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right) \). Here’s a step-by-step solution:

### Step 1: Rewrite the function
We start with the given function:
\[ y = \tan^{-1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right) \]

### Step 2: Simplify the argument of the inverse tangent
To simplify the argument, we can divide both the numerator and the denominator by \( \cos x \):
\[ y = \tan^{-1} \left( \frac{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}} \right) = \tan^{-1} \left( \frac{1 + \tan x}{1 - \tan x} \right) \]

### Step 3: Use the tangent addition formula
The expression \( \frac{1 + \tan x}{1 - \tan x} \) can be recognized as the tangent of a sum:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Setting \( A = \frac{\pi}{4} \) (where \( \tan \frac{\pi}{4} = 1 \)) and \( B = x \), we have:
\[ y = \tan^{-1}(\tan(x + \frac{\pi}{4})) \]
Since \( \tan^{-1}(\tan \theta) = \theta \) for \( \theta \) in the principal range, we get:
\[ y = x + \frac{\pi}{4} \]

### Step 4: Differentiate with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( x + \frac{\pi}{4} \right) \]
Since \( \frac{\pi}{4} \) is a constant, its derivative is 0:
\[ \frac{dy}{dx} = 1 + 0 = 1 \]

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

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