If `u=cot^(-1)sqrt(cos theta) -tan^(-1)sqrt(cos theta)` then sin u=

Options

A. `sin^(2) theta`
B. `cos^(2) theta`
C. `tan^(2) theta `
D. `tan^(2) 2theta`

Explanation

To solve the problem, we need to find \( \sin u \) where \( u = \cot^{-1}(\sqrt{\cos \theta}) - \tan^{-1}(\sqrt{\cos \theta}) \).

### Step-by-Step Solution:

1. **Express \( u \) using the identity**:
We know that:
\[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \]
Therefore, we can express \( \cot^{-1}(\sqrt{\cos \theta}) \) as:
\[ \cot^{-1}(\sqrt{\cos \theta}) = \frac{\pi}{2} - \tan^{-1}(\sqrt{\cos \theta}) \]
Substituting this into the expression for \( u \):
\[ u = \left(\frac{\pi}{2} - \tan^{-1}(\sqrt{\cos \theta})\right) - \tan^{-1}(\sqrt{\cos \theta}) \]
Simplifying this gives:
\[ u = \frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos \theta}) \]

2. **Find \( \sin u \)**:
Using the sine of a difference formula:
\[ \sin u = \sin\left(\frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos \theta})\right) \]
This simplifies to:
\[ \sin u = \cos(2\tan^{-1}(\sqrt{\cos \theta})) \]

3. **Use the double angle formula for cosine**:
The double angle formula for cosine states:
\[ \cos(2\alpha) = \frac{1 - \tan^2(\alpha)}{1 + \tan^2(\alpha)} \]
Let \( \alpha = \tan^{-1}(\sqrt{\cos \theta}) \). Then:
\[ \tan(\alpha) = \sqrt{\cos \theta} \implies \tan^2(\alpha) = \cos \theta \]
Substituting this into the double angle formula:
\[ \cos(2\tan^{-1}(\sqrt{\cos \theta})) = \frac{1 - \cos \theta}{1 + \cos \theta} \]

4. **Substitute back to find \( \sin u \)**:
Thus, we have:
\[ \sin u = \frac{1 - \cos \theta}{1 + \cos \theta} \]

5. **Use half-angle identities**:
We know that:
\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
and
\[ 1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) \]
Therefore:
\[ \sin u = \frac{2 \sin^2\left(\frac{\theta}{2}\right)}{2 \cos^2\left(\frac{\theta}{2}\right)} = \tan^2\left(\frac{\theta}{2}\right) \]

### Final Answer:
\[ \sin u = \tan^2\left(\frac{\theta}{2}\right) \]