If `alpha` is in Ist quadrant such that `sec^(2)alpha=3`, then `(tan^(2)alpha-cosec^(2)alpha)/(tan^(2)alpha+cosec^(2)alpha)=?`

Options

A. `(4)/(7)`
B. `(3)/(7)`
C. `(2)/(7)`
D. `(1)/(7)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Use the given information
We know that \( \sec^2 \alpha = 3 \). From the identity relating secant and tangent, we have:
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \]
Substituting the given value:
\[ 3 = 1 + \tan^2 \alpha \]

### Step 2: Solve for \( \tan^2 \alpha \)
Rearranging the equation gives:
\[ \tan^2 \alpha = 3 - 1 = 2 \]

### Step 3: Find \( \csc^2 \alpha \)
Using the relationship between secant and cosecant:
\[ \sec^2 \alpha = \frac{1}{\sin^2 \alpha} \implies \csc^2 \alpha = \frac{1}{\sec^2 \alpha} \]
Substituting the value of \( \sec^2 \alpha \):
\[ \csc^2 \alpha = \frac{1}{3} \]

### Step 4: Substitute values into the expression
We need to find:
\[ \frac{\tan^2 \alpha - \csc^2 \alpha}{\tan^2 \alpha + \csc^2 \alpha} \]
Substituting the values we found:
\[ \frac{2 - \frac{1}{3}}{2 + \frac{1}{3}} \]

### Step 5: Simplify the numerator and denominator
To simplify, we first find a common denominator for both the numerator and denominator:
- For the numerator:
\[ 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]
- For the denominator:
\[ 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \]

### Step 6: Final calculation
Now substituting back into the expression:
\[ \frac{\frac{5}{3}}{\frac{7}{3}} = \frac{5}{7} \]

Thus, the final answer is:
\[ \frac{5}{7} \]