If `(a^2 -b^2) sin theta` + 1ab `cos theta` = `a^2 +b^2`, then find `tan theta` .

Answer

To solve the equation \((a^2 - b^2) \sin \theta + ab \cos \theta = a^2 + b^2\) and find \(\tan \theta\), we can follow these steps:

### Step 1: Rearranging the Equation
Start by moving \(a^2 + b^2\) to the left side of the equation:
\[ (a^2 - b^2) \sin \theta + ab \cos \theta - (a^2 + b^2) = 0 \]
This can be rewritten as:
\[ (a^2 - b^2) \sin \theta + ab \cos \theta = a^2 + b^2 \]

### Step 2: Isolate Terms
Rearranging gives us:
\[ (a^2 - b^2) \sin \theta + ab \cos \theta = a^2 + b^2 \]
Now, we can express this as:
\[ \frac{(a^2 - b^2) \sin \theta + ab \cos \theta}{a^2 + b^2} = 1 \]

### Step 3: Compare with Trigonometric Identity
We know from trigonometry that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Thus, we can compare:
\[ \frac{(a^2 - b^2) \sin \theta}{a^2 + b^2} + \frac{ab \cos \theta}{a^2 + b^2} = 1 \]

### Step 4: Assign Values to Sin and Cos
From the comparison, we can assign:
\[ \sin \theta = \frac{a^2 - b^2}{a^2 + b^2} \]
\[ \cos \theta = \frac{ab}{a^2 + b^2} \]

### Step 5: Calculate \(\tan \theta\)
Now, we can find \(\tan \theta\) using the definition:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting the values we found:
\[ \tan \theta = \frac{\frac{a^2 - b^2}{a^2 + b^2}}{\frac{ab}{a^2 + b^2}} \]

### Step 6: Simplify
This simplifies to:
\[ \tan \theta = \frac{a^2 - b^2}{ab} \]

### Final Answer
Thus, the value of \(\tan \theta\) is:
\[ \tan \theta = \frac{a^2 - b^2}{ab} \]