if `cot^(-1)[sqrt(cosalpha)]-tan^(-1)[sqrt(cosalpha)]=x` then `sinx` is equal to

Options

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

Explanation

To solve the problem, we need to find \( \sin x \) given that

\[ \cot^{-1}(\sqrt{\cos \alpha}) - \tan^{-1}(\sqrt{\cos \alpha}) = x. \]

### Step 1: Rewrite the expression using the tangent function

We can rewrite \( \cot^{-1}(y) \) in terms of \( \tan^{-1}(y) \):

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

Thus, we have:

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

Substituting this into our equation gives:

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

### Step 2: Simplify the expression

This simplifies to:

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

### Step 3: Solve for \( \tan^{-1}(\sqrt{\cos \alpha}) \)

Rearranging gives:

\[ \tan^{-1}(\sqrt{\cos \alpha}) = \frac{\pi}{4} - \frac{x}{2}. \]

### Step 4: Find \( \tan x \)

Taking the tangent of both sides, we have:

\[ \tan\left(\tan^{-1}(\sqrt{\cos \alpha})\right) = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right). \]

This leads to:

\[ \sqrt{\cos \alpha} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right). \]

Using the tangent subtraction formula:

\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}, \]

where \( A = \frac{\pi}{4} \) and \( B = \frac{x}{2} \), we have:

\[ \tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \tan\left(\frac{x}{2}\right) = t. \]

Substituting these into the formula gives:

\[ \sqrt{\cos \alpha} = \frac{1 - t}{1 + t}. \]

### Step 5: Solve for \( t \)

Cross-multiplying yields:

\[ \sqrt{\cos \alpha}(1 + t) = 1 - t. \]

Rearranging gives:

\[ \sqrt{\cos \alpha} + \sqrt{\cos \alpha} t = 1 - t. \]

Combining like terms results in:

\[ t(\sqrt{\cos \alpha} + 1) = 1 - \sqrt{\cos \alpha}. \]

Thus,

\[ t = \frac{1 - \sqrt{\cos \alpha}}{1 + \sqrt{\cos \alpha}}. \]

### Step 6: Find \( \sin x \)

Now, we know:

\[ \tan\left(\frac{x}{2}\right) = t = \frac{1 - \sqrt{\cos \alpha}}{1 + \sqrt{\cos \alpha}}. \]

Using the identity \( \sin x = \frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \):

Let \( \tan\left(\frac{x}{2}\right) = t \):

\[ \sin x = \frac{2t}{1 + t^2}. \]

Substituting \( t \):

\[ \sin x = \frac{2\left(\frac{1 - \sqrt{\cos \alpha}}{1 + \sqrt{\cos \alpha}}\right)}{1 + \left(\frac{1 - \sqrt{\cos \alpha}}{1 + \sqrt{\cos \alpha}}\right)^2}. \]

### Step 7: Simplify the expression

After simplification, using the half-angle formulas:

\[ \sin x = \frac{2 \sin^2\left(\frac{\alpha}{2}\right)}{2 \cos^2\left(\frac{\alpha}{2}\right)} = \tan^2\left(\frac{\alpha}{2}\right). \]

Thus, we conclude that:

\[ \sin x = \tan^2\left(\frac{\alpha}{2}\right). \]

### Final Answer

The final answer is:

\[ \sin x = \tan^2\left(\frac{\alpha}{2}\right). \]