`(d)/(dx)[tan^(-1)((asinx+bcosx)/(acosx-bsinx))]=`

Options

A. `(a)/(b)`
B. `(b)/(a)`
C. 1
D. 0

Answer

To solve the problem \(\frac{d}{dx}\left[\tan^{-1}\left(\frac{a \sin x + b \cos x}{a \cos x - b \sin x}\right)\right]\), we will follow these steps:

### Step 1: Simplify the Argument of the Arctangent
We start by simplifying the expression inside the arctangent. We divide both the numerator and the denominator by \(a \cos x\):

\[ \frac{a \sin x + b \cos x}{a \cos x - b \sin x} = \frac{\frac{a \sin x}{a \cos x} + \frac{b \cos x}{a \cos x}}{\frac{a \cos x}{a \cos x} - \frac{b \sin x}{a \cos x}} = \frac{\tan x + \frac{b}{a}}{1 - \frac{b}{a} \tan x} \]

### Step 2: Use the Arctangent Addition Formula
Now, we can use the identity for the tangent of a sum:

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

where \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\).

### Step 3: Apply the Inverse Tangent Function
Since \(\tan^{-1}\) and \(\tan\) are inverse functions, we have:

\[ \tan^{-1}(\tan(x + \theta)) = x + \theta \]

### Step 4: Differentiate the Result
Now we differentiate \(x + \theta\) with respect to \(x\):

\[ \frac{d}{dx}(x + \theta) = \frac{d}{dx}(x) + \frac{d}{dx}(\theta) \]

Since \(\theta\) is a constant (as it depends on \(a\) and \(b\)), its derivative is 0:

\[ \frac{d}{dx}(x + \theta) = 1 + 0 = 1 \]

### Final Answer
Thus, the derivative is:

\[ \frac{d}{dx}\left[\tan^{-1}\left(\frac{a \sin x + b \cos x}{a \cos x - b \sin x}\right)\right] = 1 \]

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