If `cot^(-1) sqrt( cos alpha)- tan^(-1) sqrt(cos alpha) =x`, then sin x equals

Options

A. `tan^(2)""((alpha)/(2))`
B. `cot^(2) ((alpha)/(2))`
C. `tan alpha`
D. `cot ((alpha)/(2))`

Answer

To solve the equation \( \cot^{-1}(\sqrt{\cos \alpha}) - \tan^{-1}(\sqrt{\cos \alpha}) = x \) and find \( \sin x \), we can follow these steps:

### Step 1: Use the identity for \( \cot^{-1} \) and \( \tan^{-1} \)
We know that:
\[ \cot^{-1}(y) - \tan^{-1}(y) = \frac{\pi}{2} - 2\tan^{-1}(y) \]
Thus, we can rewrite the equation as:
\[ \frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos \alpha}) = x \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ x = \frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos \alpha}) \]

### Step 3: Finding \( \sin x \)
Using the sine of a difference identity, we have:
\[ \sin x = \sin\left(\frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos \alpha})\right) = \cos(2\tan^{-1}(\sqrt{\cos \alpha})) \]

### Step 4: Use the double angle formula
Using the double angle formula for cosine:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Let \( \theta = \tan^{-1}(\sqrt{\cos \alpha}) \). Then:
\[ \sin(\theta) = \frac{\sqrt{\cos \alpha}}{\sqrt{1 + \cos \alpha}} \]
and
\[ \cos(\theta) = \frac{1}{\sqrt{1 + \cos \alpha}} \]

### Step 5: Calculate \( \sin^2(\theta) \)
Now we can find \( \sin^2(\theta) \):
\[ \sin^2(\theta) = \frac{\cos \alpha}{1 + \cos \alpha} \]

### Step 6: Substitute back into the cosine formula
Now substituting back into the double angle formula:
\[ \cos(2\tan^{-1}(\sqrt{\cos \alpha})) = 1 - 2\left(\frac{\cos \alpha}{1 + \cos \alpha}\right) \]
Simplifying gives:
\[ \cos(2\tan^{-1}(\sqrt{\cos \alpha})) = \frac{1 + \cos \alpha - 2\cos \alpha}{1 + \cos \alpha} = \frac{1 - \cos \alpha}{1 + \cos \alpha} \]

### Final Result
Thus, we find that:
\[ \sin x = \frac{1 - \cos \alpha}{1 + \cos \alpha} \]