Find the quadrants of the coordinate planes such that for each point `(x,y)` on these quadrants ( where ` x ne 0 , y ne 0`) , the equation, ` (sin^(4) theta )/x + ( cos^4 theta)/( y)=(1)/(x+y) ` is soluble for ` theta ` .

Answer

To solve the problem, we need to analyze the equation given:

\[ \frac{\sin^4 \theta}{x} + \frac{\cos^4 \theta}{y} = \frac{1}{x+y} \]

### Step 1: Rewrite the equation
We start by rewriting the equation to make it easier to manipulate. We can take the least common multiple (LCM) of the left-hand side:

\[ \frac{y \sin^4 \theta + x \cos^4 \theta}{xy} = \frac{1}{x+y} \]

### Step 2: Cross-multiply
Next, we cross-multiply to eliminate the fractions:

\[ (x+y)(y \sin^4 \theta + x \cos^4 \theta) = xy \]

### Step 3: Expand and rearrange
Expanding the left-hand side gives us:

\[ y(x+y) \sin^4 \theta + x(x+y) \cos^4 \theta = xy \]

Rearranging this, we have:

\[ y(x+y) \sin^4 \theta + x(x+y) \cos^4 \theta - xy = 0 \]

### Step 4: Factor out common terms
We can factor out \(xy\) from the equation:

\[ y(x+y) \sin^4 \theta + x(x+y) \cos^4 \theta = xy \]

### Step 5: Analyze the equation
Now we need to analyze the conditions under which this equation is solvable for \(\theta\).

We can express \(\sin^4 \theta\) and \(\cos^4 \theta\) in terms of \(\sin^2 \theta\) and \(\cos^2 \theta\):

\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 6: Set up the condition
For the equation to be solvable, the term \(x \cos^2 \theta - y \sin^2 \theta\) must equal zero:

\[ x \cos^2 \theta = y \sin^2 \theta \]

This implies:

\[ \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{x}{y} \implies \tan^2 \theta = \frac{x}{y} \]

### Step 7: Determine the quadrants
The expression \(\tan^2 \theta\) is always positive. Therefore, \(\frac{x}{y}\) must also be positive. This occurs in two cases:

1. \(x > 0\) and \(y > 0\) (First Quadrant)
2. \(x < 0\) and \(y < 0\) (Third Quadrant)

### Conclusion
Thus, the quadrants of the coordinate plane where the equation is soluble for \(\theta\) are:

**First Quadrant** and **Third Quadrant**.