Let `u = cot^(-1) sqrt(cos 2 theta) - tan^(-1) sqrt( cos 2 theta)` , then the value of `sin u` is

Options

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

Answer

To solve the problem, we need to find the value of \( \sin u \) where

\[ u = \cot^{-1}(\sqrt{\cos 2\theta}) - \tan^{-1}(\sqrt{\cos 2\theta}). \]

### Step 1: Rewrite \( u \)

Using the identity \( \cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2} \), we can rewrite \( u \):

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

### Step 2: Substitute \( x \)

Let \( x = \tan^{-1}(\sqrt{\cos 2\theta}) \). Then we can express \( u \) as:

\[ u = \frac{\pi}{2} - 2x. \]

### Step 3: Find \( \sin u \)

Using the sine of a difference identity, we have:

\[ \sin u = \sin\left(\frac{\pi}{2} - 2x\right) = \cos(2x). \]

### Step 4: Express \( \cos(2x) \)

Using the double angle formula for cosine, we can express \( \cos(2x) \) in terms of \( \tan x \):

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

### Step 5: Substitute \( \tan x \)

Since \( x = \tan^{-1}(\sqrt{\cos 2\theta}) \), we have:

\[ \tan x = \sqrt{\cos 2\theta}. \]

### Step 6: Substitute into \( \cos(2x) \)

Now, substituting \( \tan x \) into the expression for \( \cos(2x) \):

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

### Step 7: Simplify the expression

Using the identity \( 1 - \cos 2\theta = 2\sin^2 \theta \) and \( 1 + \cos 2\theta = 2\cos^2 \theta \), we can simplify:

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

### Conclusion

Thus, the value of \( \sin u \) is:

\[ \sin u = \tan^2 \theta. \]

### Final Answer

The value of \( \sin u \) is \( \tan^2 \theta \).