If `tantheta=sqrt((1)/(13))and theta` is acute , then what is the value of `((cosec^(2)theta-sec^(2)theta))/(cosec^(2)theta+sec^(2)theta))` ?

Options

A. `(2)/(3)`
B. `(3)/(4)`
C. `(5)/(6)`
D. `(6)/(7)`

Answer

To solve the problem, we need to find the value of \(\frac{\csc^2 \theta - \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta}\) given that \(\tan \theta = \sqrt{\frac{1}{13}}\) and \(\theta\) is acute.

### Step-by-Step Solution:

1. **Given Information**:
\[ \tan \theta = \sqrt{\frac{1}{13}} \implies \tan \theta = \frac{1}{\sqrt{13}} \]

2. **Finding \(\sin \theta\) and \(\cos \theta\)**:
From the definition of tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{13}} \]
We can represent \(\sin \theta\) and \(\cos \theta\) in terms of a right triangle. Let the opposite side be \(1\) and the adjacent side be \(\sqrt{13}\). The hypotenuse \(h\) can be calculated using the Pythagorean theorem:
\[ h = \sqrt{1^2 + (\sqrt{13})^2} = \sqrt{1 + 13} = \sqrt{14} \]
Now we can find \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = \frac{1}{\sqrt{14}}, \quad \cos \theta = \frac{\sqrt{13}}{\sqrt{14}} \]

3. **Finding \(\csc^2 \theta\) and \(\sec^2 \theta\)**:
Using the definitions:
\[ \csc^2 \theta = \frac{1}{\sin^2 \theta} = \frac{1}{\left(\frac{1}{\sqrt{14}}\right)^2} = 14 \]
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{1}{\left(\frac{\sqrt{13}}{\sqrt{14}}\right)^2} = \frac{14}{13} \]

4. **Substituting into the expression**:
Now substitute \(\csc^2 \theta\) and \(\sec^2 \theta\) into the expression:
\[ \frac{\csc^2 \theta - \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta} = \frac{14 - \frac{14}{13}}{14 + \frac{14}{13}} \]

5. **Simplifying the numerator**:
To simplify the numerator:
\[ 14 - \frac{14}{13} = \frac{14 \cdot 13}{13} - \frac{14}{13} = \frac{182 - 14}{13} = \frac{168}{13} \]

6. **Simplifying the denominator**:
To simplify the denominator:
\[ 14 + \frac{14}{13} = \frac{14 \cdot 13}{13} + \frac{14}{13} = \frac{182 + 14}{13} = \frac{196}{13} \]

7. **Final Calculation**:
Now substitute back into the expression:
\[ \frac{\frac{168}{13}}{\frac{196}{13}} = \frac{168}{196} = \frac{6}{7} \]

Thus, the final answer is:
\[ \frac{6}{7} \]