If `(pi)/(2) lt theta lt (3pi)/(2)` then `sqrt(tan^(2)theta-sin^(2)theta)` is equal to :

Options

A. `tan theta sin theta`
B. `-tan theta sin theta`
C. `tan theta-sin theta`
D. `sin theta-tan theta`

Answer

To solve the problem, we need to find the value of the expression \(\sqrt{\tan^2 \theta - \sin^2 \theta}\) given that \(\frac{\pi}{2} < \theta < \frac{3\pi}{2}\).

### Step-by-Step Solution:

1. **Start with the expression:**
\[ \sqrt{\tan^2 \theta - \sin^2 \theta} \]

2. **Rewrite \(\tan^2 \theta\):**
Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), thus:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Substitute this into the expression:
\[ \sqrt{\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta} \]

3. **Combine the terms under a common denominator:**
\[ \sqrt{\frac{\sin^2 \theta - \sin^2 \theta \cos^2 \theta}{\cos^2 \theta}} \]
Factor out \(\sin^2 \theta\):
\[ \sqrt{\frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta}} \]

4. **Use the Pythagorean identity:**
We know that \(1 - \cos^2 \theta = \sin^2 \theta\):
\[ \sqrt{\frac{\sin^2 \theta \cdot \sin^2 \theta}{\cos^2 \theta}} = \sqrt{\frac{\sin^4 \theta}{\cos^2 \theta}} \]

5. **Simplify the square root:**
\[ = \frac{\sin^2 \theta}{|\cos \theta|} \]

6. **Determine the sign of \(\cos \theta\):**
Since \(\frac{\pi}{2} < \theta < \frac{3\pi}{2}\), \(\cos \theta\) is negative in both the second and third quadrants. Therefore, \(|\cos \theta| = -\cos \theta\).

7. **Substitute back:**
\[ = \frac{\sin^2 \theta}{-\cos \theta} = -\frac{\sin^2 \theta}{\cos \theta} \]

8. **Final expression:**
\[ = -\tan \theta \cdot \sin \theta \]

### Conclusion:
Thus, the value of \(\sqrt{\tan^2 \theta - \sin^2 \theta}\) is:
\[ -\tan \theta \cdot \sin \theta \]