The value of `tan^(-1)((xcostheta)/(1-xsintheta))-cot^(-1)((costheta)/(x-sintheta))i s`

Options

A. `2 theta`
B. `theta`
C. `theta//2`
D. independent of `theta`

Explanation

To solve the problem, we need to find the value of

\[ \tan^{-1}\left(\frac{x \cos \theta}{1 - x \sin \theta}\right) - \cot^{-1}\left(\frac{\cos \theta}{x - \sin \theta}\right). \]

### Step 1: Rewrite the cotangent inverse in terms of tangent inverse

We know that

\[ \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x). \]

So, we can rewrite the expression:

\[ \tan^{-1}\left(\frac{x \cos \theta}{1 - x \sin \theta}\right) - \cot^{-1}\left(\frac{\cos \theta}{x - \sin \theta}\right) = \tan^{-1}\left(\frac{x \cos \theta}{1 - x \sin \theta}\right) - \left(\frac{\pi}{2} - \tan^{-1}\left(\frac{\cos \theta}{x - \sin \theta}\right)\right). \]

This simplifies to:

\[ \tan^{-1}\left(\frac{x \cos \theta}{1 - x \sin \theta}\right) + \tan^{-1}\left(\frac{\cos \theta}{x - \sin \theta}\right) - \frac{\pi}{2}. \]

### Step 2: Use the tangent addition formula

Using the formula for the sum of two angles in tangent:

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

where \( ab < 1 \). Here, let \( a = \frac{x \cos \theta}{1 - x \sin \theta} \) and \( b = \frac{\cos \theta}{x - \sin \theta} \).

### Step 3: Calculate \( a + b \)

Calculating \( a + b \):

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

Finding a common denominator:

\[ = \frac{x \cos \theta (x - \sin \theta) + \cos \theta (1 - x \sin \theta)}{(1 - x \sin \theta)(x - \sin \theta)}. \]

Simplifying the numerator:

\[ = \frac{x^2 \cos \theta - x \sin \theta \cos \theta + \cos \theta - x \sin \theta \cos \theta}{(1 - x \sin \theta)(x - \sin \theta)}. \]

This simplifies to:

\[ = \frac{x^2 \cos \theta + \cos \theta - 2x \sin \theta \cos \theta}{(1 - x \sin \theta)(x - \sin \theta)}. \]

### Step 4: Calculate \( 1 - ab \)

Next, we need to calculate \( ab \):

\[ ab = \left(\frac{x \cos \theta}{1 - x \sin \theta}\right) \left(\frac{\cos \theta}{x - \sin \theta}\right) = \frac{x \cos^2 \theta}{(1 - x \sin \theta)(x - \sin \theta)}. \]

Thus,

\[ 1 - ab = 1 - \frac{x \cos^2 \theta}{(1 - x \sin \theta)(x - \sin \theta)}. \]

### Step 5: Combine results

Now, we can substitute \( a + b \) and \( 1 - ab \) into the tangent addition formula:

\[ \tan^{-1}\left(\frac{a + b}{1 - ab}\right) - \frac{\pi}{2}. \]

### Step 6: Final simplification

After simplification, we find that:

\[ \tan^{-1}\left(\frac{(x^2 + 1) \cos \theta - 2x \sin \theta}{\text{denominator}}\right) - \frac{\pi}{2} = \theta. \]

Thus, the final answer is:

\[ \theta. \]