If `(a^(2)-b^(2))sintheta+2abcostheta=a^(2)+b^(2)` ,then `tantheta=`?

Options

A. `(2ab)/(a^(2)-b^(2))`
B. `(a^(2)-b^(2))/(2ab)`
C. `(ab)/(a^(2)-b^(2))`
D. `(a^(2)-b^(2))/(ab)`

Answer

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

### Step 1: Identify the components
From the equation, we can identify:
- \(A = a^2 - b^2\) (the coefficient of \(\sin \theta\))
- \(B = 2ab\) (the coefficient of \(\cos \theta\))
- \(C = a^2 + b^2\) (the constant on the right side)

### Step 2: Check the Pythagorean identity
We need to check if \(A^2 + B^2 = C^2\):
\[ A^2 = (a^2 - b^2)^2 = a^4 - 2a^2b^2 + b^4 \]
\[ B^2 = (2ab)^2 = 4a^2b^2 \]
Now, adding these:
\[ A^2 + B^2 = (a^4 - 2a^2b^2 + b^4) + 4a^2b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2 \]
Now, calculate \(C^2\):
\[ C^2 = (a^2 + b^2)^2 \]
Since \(A^2 + B^2 = C^2\), we can use the triangle representation.

### Step 3: Form the triangle
In the triangle:
- The height (perpendicular) is \(A = a^2 - b^2\).
- The base is \(B = 2ab\).
- The hypotenuse is \(C = a^2 + b^2\).

### Step 4: Find \(\tan \theta\)
Using the definition of tangent in a right triangle:
\[ \tan \theta = \frac{\text{perpendicular}}{\text{base}} = \frac{A}{B} = \frac{a^2 - b^2}{2ab} \]

### Final Answer
Thus, we have:
\[ \tan \theta = \frac{a^2 - b^2}{2ab} \]

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