Let:
Equation Setup
Let:
\[ x = \tan^2 \theta \]
\[ y = \sec^2 \theta \]
1) \(\tan \theta + \sec^2 \theta = 7\)
2) \(\tan^2 \theta + \sec \theta = 11\)
Solving Equation 1
Take 1):
\[ \tan \theta + \frac{1}{\cos^2 \theta} = 7 \]
Solve it, and you will get:
\[ 7\sin^2 \theta + \sin \theta - 6 = 0 \quad (A) \]
Solving Equation 2
Now, similarly, take equation (2) and solve it.
You will get:
\[ 12\cos^2 \theta - \cos \theta - 1 = 0 \quad (B) \]
Solving Equation A
Taking equation (A):
Let:
\[ \sin \theta = t \]
\[ 7t^2 + t - 6 = 0 \]
Solve it, and you will get:
\[ t = -1 \text{ or } t = \frac{6}{7} \]
\[ \sin \theta = -1 \text{ or } \sin \theta = \frac{6}{7} \]
Solving Equation B
Now, take equation (B) and do similarly as equation A:
\[ 12\cos^2 \theta - \cos \theta - 1 = 0 \]
Solve it as above:
\[ \cos \theta = \frac{1}{3} \text{ or } \cos \theta = -\frac{1}{4} \]
Since the cosine value exists in the 3rd quadrant, not negative, \(-\frac{1}{4}\) is not possible. Therefore:
\[ \cos \theta = \frac{1}{3} \]
The sine value exists in the 3rd quadrant, not positive, so:
\[ \sin \theta = -1 \]
Final Calculation
Now,
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = -1/\frac{1}{3} = -3 \]
\[ x = \tan^2 \theta = (-3)^2 = 9 \]
Put in equation 1:
\[ y = 4 \]
Therefore:
\[ x = 9 \text{ and } y = 4 \text{ is the answer.} \]