If x = `a(sintheta + costheta) and y = b (sintheta- costheta)` then `(x^2 //a^2 ) + (y^2 //b^2) =?`
(a)3
(b)1
(c)4
(d)2

Options

A. 3
B. 1
C. 4
D. 2

Answer

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

1. \( x = a(\sin \theta + \cos \theta) \)
2. \( y = b(\sin \theta - \cos \theta) \)

We need to find the value of:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \]

### Step 1: Substitute the expressions for \(x\) and \(y\)

Substituting the values of \(x\) and \(y\) into the expression:

\[ \frac{x^2}{a^2} = \frac{(a(\sin \theta + \cos \theta))^2}{a^2} = (\sin \theta + \cos \theta)^2 \]

\[ \frac{y^2}{b^2} = \frac{(b(\sin \theta - \cos \theta))^2}{b^2} = (\sin \theta - \cos \theta)^2 \]

### Step 2: Expand the squares

Now we expand both squares:

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

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

### Step 3: Combine the results

Now we add these two results together:

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

Combining like terms gives:

\[ 2\sin^2 \theta + 2\cos^2 \theta \]

### Step 4: Use the Pythagorean identity

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

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

### Final Result

Thus, we find that:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \]

The answer is **(d) 2**.

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