If `a+btantheta=secthetaand b-atantheta=3sectheta`, then findthe value of `a^2+b^2`.

Explanation

To solve the problem, we start with the given equations:

1. \( a + b \tan \theta = \sec \theta \)
2. \( b - a \tan \theta = 3 \sec \theta \)

### Step 1: Rewrite the equations in terms of sine and cosine.

We know that:
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
- \( \sec \theta = \frac{1}{\cos \theta} \)

Substituting these into the equations, we have:

1. \( a + b \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \)
2. \( b - a \frac{\sin \theta}{\cos \theta} = 3 \frac{1}{\cos \theta} \)

### Step 2: Clear the denominators.

Multiply both sides of each equation by \( \cos \theta \):

1. \( a \cos \theta + b \sin \theta = 1 \) (Equation 1)
2. \( b \cos \theta - a \sin \theta = 3 \) (Equation 2)

### Step 3: Square both equations and add them.

Now we will square both equations and add them:

1. \( (a \cos \theta + b \sin \theta)^2 = 1^2 \)
2. \( (b \cos \theta - a \sin \theta)^2 = 3^2 \)

Expanding both equations:

**From Equation 1:**
\[ (a \cos \theta + b \sin \theta)^2 = a^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta = 1 \]

**From Equation 2:**
\[ (b \cos \theta - a \sin \theta)^2 = b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta + a^2 \sin^2 \theta = 9 \]

### Step 4: Add the two expanded equations.

Now we add the two results:

\[ (a^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta) + (b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta + a^2 \sin^2 \theta) = 1 + 9 \]

This simplifies to:

\[ a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = 10 \]

### Step 5: Use the Pythagorean identity.

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we can simplify:

\[ a^2 \cdot 1 + b^2 \cdot 1 = 10 \]

Thus, we have:

\[ a^2 + b^2 = 10 \]

### Final Answer:

The value of \( a^2 + b^2 \) is \( \boxed{10} \).