If `sintheta,tantheta,costheta` are in G.P. then `4sin^2theta-3sin^4theta+sin^6theta=_________`

Options

A. `-1`
B. 2
C. 1
D. none of these

Explanation

To solve the problem, we need to find the value of \(4\sin^2\theta - 3\sin^4\theta + \sin^6\theta\) given that \(\sin\theta\), \(\tan\theta\), and \(\cos\theta\) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the G.P. Condition**:
Since \(\sin\theta\), \(\tan\theta\), and \(\cos\theta\) are in G.P., we can use the property of G.P. which states that if \(a\), \(b\), and \(c\) are in G.P., then \(b^2 = ac\). Here, we have:
\[ \tan^2\theta = \sin\theta \cdot \cos\theta \]

2. **Expressing \(\tan\theta\)**:
Recall that \(\tan\theta = \frac{\sin\theta}{\cos\theta}\). Therefore:
\[ \tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta} \]
Substituting this into the G.P. condition gives:
\[ \frac{\sin^2\theta}{\cos^2\theta} = \sin\theta \cdot \cos\theta \]

3. **Cross Multiplying**:
Cross-multiplying yields:
\[ \sin^2\theta = \sin\theta \cdot \cos^3\theta \]

4. **Dividing by \(\sin\theta\)** (assuming \(\sin\theta \neq 0\)):
Dividing both sides by \(\sin\theta\) gives:
\[ \sin\theta = \cos^3\theta \]

5. **Substituting into the Expression**:
Now, we need to substitute \(\sin\theta = \cos^3\theta\) into the expression \(4\sin^2\theta - 3\sin^4\theta + \sin^6\theta\):
- First, calculate \(\sin^2\theta\):
\[ \sin^2\theta = (\cos^3\theta)^2 = \cos^6\theta \]
- Next, calculate \(\sin^4\theta\):
\[ \sin^4\theta = (\cos^3\theta)^4 = \cos^{12}\theta \]
- Finally, calculate \(\sin^6\theta\):
\[ \sin^6\theta = (\cos^3\theta)^6 = \cos^{18}\theta \]

6. **Substituting Values**:
Now substitute these into the original expression:
\[ 4\sin^2\theta - 3\sin^4\theta + \sin^6\theta = 4\cos^6\theta - 3\cos^{12}\theta + \cos^{18}\theta \]

7. **Factoring the Expression**:
We can factor out \(\cos^6\theta\):
\[ = \cos^6\theta (4 - 3\cos^6\theta + \cos^{12}\theta) \]

8. **Using the Identity**:
Notice that \(1 - \sin^2\theta = \cos^2\theta\), so we can express \(\cos^6\theta\) in terms of \(\sin^2\theta\):
\[ \cos^6\theta = (1 - \sin^2\theta)^3 \]

9. **Final Calculation**:
Since we know that \(\sin^2\theta + \cos^2\theta = 1\), we can substitute and simplify:
\[ 4 - 3\sin^2\theta + \sin^4\theta = 1 \]

Thus, the final value is:
\[ \boxed{1} \]