If `0^(@)ltthetalt90^(@)and"cosectheta=cot^(2)theta` then the value of the expression `"cos"^(4)theta+2"cos"^(6)theta+cos^(8)theta` is equal to :

Options

A. 2
B. 0
C. 1
D. 3

Answer

To solve the problem, we need to find the value of the expression \( \cos^4 \theta + 2 \cos^6 \theta + \cos^8 \theta \) given that \( \csc \theta = \cot^2 \theta \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \csc \theta = \cot^2 \theta \]
Recall that \( \csc \theta = \frac{1}{\sin \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), so we can rewrite \( \cot^2 \theta \) as:
\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \]
Therefore, the equation becomes:
\[ \frac{1}{\sin \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \]

2. **Cross-multiply to eliminate the fractions**:
\[ \sin^2 \theta = \cos^2 \theta \]
This implies:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Since \( \sin^2 \theta = \cos^2 \theta \), we can set:
\[ 2\cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{2} \]

3. **Find \( \cos \theta \)**:
Since \( \cos^2 \theta = \frac{1}{2} \), we have:
\[ \cos \theta = \frac{1}{\sqrt{2}} \quad \text{(since } 0 < \theta < 90^\circ\text{)} \]

4. **Calculate \( \cos^4 \theta \), \( \cos^6 \theta \), and \( \cos^8 \theta \)**:
\[ \cos^4 \theta = \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} \]
\[ \cos^6 \theta = \left(\frac{1}{\sqrt{2}}\right)^6 = \frac{1}{8} \]
\[ \cos^8 \theta = \left(\frac{1}{\sqrt{2}}\right)^8 = \frac{1}{16} \]

5. **Substitute these values into the expression**:
\[ \cos^4 \theta + 2 \cos^6 \theta + \cos^8 \theta = \frac{1}{4} + 2 \cdot \frac{1}{8} + \frac{1}{16} \]
Simplifying the terms:
\[ = \frac{1}{4} + \frac{2}{8} + \frac{1}{16} \]
\[ = \frac{1}{4} + \frac{1}{4} + \frac{1}{16} \]
Convert \( \frac{1}{4} \) to sixteenths:
\[ = \frac{4}{16} + \frac{4}{16} + \frac{1}{16} = \frac{9}{16} \]

6. **Final Result**:
Therefore, the value of the expression \( \cos^4 \theta + 2 \cos^6 \theta + \cos^8 \theta \) is:
\[ \frac{9}{16} \]