If `(cosec theta + cot theta) //(cosec theta - cot theta) = 7` , the find the value of `(4sin^2 theta - 1)//(4 sin^2 theta + 5)`

Options

A. `2//9`
B. `1//9`
C. `1//3`
D. `4//9`

Answer

To solve the problem, we need to find the value of \(\frac{4\sin^2 \theta - 1}{4\sin^2 \theta + 5}\) given that \(\frac{\csc \theta + \cot \theta}{\csc \theta - \cot \theta} = 7\).

### Step-by-Step Solution:

1. **Rewrite the given equation**:
\[ \frac{\csc \theta + \cot \theta}{\csc \theta - \cot \theta} = 7 \]
We can express \(\csc \theta\) and \(\cot \theta\) in terms of \(\sin \theta\) and \(\cos \theta\):
\[ \csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Thus, we can rewrite the equation as:
\[ \frac{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}} = 7 \]

2. **Simplify the equation**:
The equation simplifies to:
\[ \frac{1 + \cos \theta}{1 - \cos \theta} = 7 \]

3. **Cross-multiply to eliminate the fraction**:
\[ 1 + \cos \theta = 7(1 - \cos \theta) \]
Expanding the right side gives:
\[ 1 + \cos \theta = 7 - 7\cos \theta \]

4. **Rearrange the equation**:
Bringing all terms involving \(\cos \theta\) to one side:
\[ \cos \theta + 7\cos \theta = 7 - 1 \]
This simplifies to:
\[ 8\cos \theta = 6 \]
Therefore:
\[ \cos \theta = \frac{6}{8} = \frac{3}{4} \]

5. **Find \(\sin^2 \theta\)** using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \(\cos^2 \theta\):
\[ \sin^2 \theta + \left(\frac{3}{4}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{9}{16} = 1 \]
\[ \sin^2 \theta = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \]

6. **Substitute \(\sin^2 \theta\) into the expression**:
Now we need to find:
\[ \frac{4\sin^2 \theta - 1}{4\sin^2 \theta + 5} \]
Substituting \(\sin^2 \theta = \frac{7}{16}\):
\[ 4\sin^2 \theta = 4 \times \frac{7}{16} = \frac{28}{16} = \frac{7}{4} \]
Now substituting into the expression:
\[ \frac{\frac{7}{4} - 1}{\frac{7}{4} + 5} = \frac{\frac{7}{4} - \frac{4}{4}}{\frac{7}{4} + \frac{20}{4}} = \frac{\frac{3}{4}}{\frac{27}{4}} = \frac{3}{27} = \frac{1}{9} \]

### Final Answer:
\[ \frac{4\sin^2 \theta - 1}{4\sin^2 \theta + 5} = \frac{1}{9} \]