Let x and y be 2 real numbers which satisfy the equations `(tan^(2)x - sec^(2)y )= (5a)/(6) - 3 ` and ` (- sec^(2)x + tan^(2)y ) = a^(2) ` , then the product of all possible value's of a can be equal to :

Options

A. `0`
B. ` -(2)/(3)`
C. `-1`
D. ` -(3)/(2)`

Answer

To solve the problem, we need to find the product of all possible values of \( a \) given the equations:

1. \( \tan^2 x - \sec^2 y = \frac{5a}{6} - 3 \)
2. \( -\sec^2 x + \tan^2 y = a^2 \)

### Step 1: Rewrite the equations
Let's rewrite the equations for clarity:

- Equation (1):
\[ \tan^2 x - \sec^2 y = \frac{5a}{6} - 3 \]

- Equation (2):
\[ -\sec^2 x + \tan^2 y = a^2 \]

### Step 2: Add the two equations
Now, we will add both equations:

\[ (\tan^2 x - \sec^2 y) + (-\sec^2 x + \tan^2 y) = \left(\frac{5a}{6} - 3\right) + a^2 \]

This simplifies to:

\[ \tan^2 x + \tan^2 y - \sec^2 x - \sec^2 y = \frac{5a}{6} - 3 + a^2 \]

### Step 3: Use the identity
We know from trigonometric identities that:

\[ \tan^2 \theta - \sec^2 \theta = -1 \]

Thus, we can rewrite the left-hand side:

\[ (\tan^2 x - \sec^2 x) + (\tan^2 y - \sec^2 y) = -1 - 1 = -2 \]

So we have:

\[ -2 = \frac{5a}{6} - 3 + a^2 \]

### Step 4: Rearrange the equation
Rearranging gives:

\[ a^2 + \frac{5a}{6} - 1 = 0 \]

### Step 5: Eliminate the fraction
To eliminate the fraction, multiply the entire equation by 6:

\[ 6a^2 + 5a - 6 = 0 \]

### Step 6: Use the quadratic formula
Now, we can use the quadratic formula to find the values of \( a \):

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 6 \), \( b = 5 \), and \( c = -6 \):

\[ a = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-6)}}{2 \cdot 6} \]
\[ = \frac{-5 \pm \sqrt{25 + 144}}{12} \]
\[ = \frac{-5 \pm \sqrt{169}}{12} \]
\[ = \frac{-5 \pm 13}{12} \]

### Step 7: Calculate the two possible values of \( a \)
Calculating the two possible values:

1. \( a = \frac{-5 + 13}{12} = \frac{8}{12} = \frac{2}{3} \)
2. \( a = \frac{-5 - 13}{12} = \frac{-18}{12} = -\frac{3}{2} \)

### Step 8: Find the product of the values
Now, we need to find the product of these two values:

\[ \text{Product} = \left(-\frac{3}{2}\right) \cdot \left(\frac{2}{3}\right) = -1 \]

### Final Answer
Thus, the product of all possible values of \( a \) is:

\[ \boxed{-1} \]