If `cos ^(-1)\ (x)/(a) + cos^(-1)\ (y)/(b) = theta`, then `(x^(2))/(a^(2)) +(y^(2))/(b^(2))=?`

Options

A. `(xy)/(ab) cos theta + cos^(2) theta`
B. `(2xy)/(ab) cos theta + cos^(2) theta`
C. `(2xy)/(ab) cos theta + sin^(2) theta`
D. `(xy)/(ab) cos theta + sin^(2) theta`

Answer

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

\[ \cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \theta \]

### Step 1: Use the formula for the sum of inverse cosines
We can use the identity for the sum of two inverse cosines:

\[ \cos^{-1}(u) + \cos^{-1}(v) = \cos^{-1}(uv - \sqrt{(1-u^2)(1-v^2)}) \]

Here, let \( u = \frac{x}{a} \) and \( v = \frac{y}{b} \). Thus, we have:

\[ \cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \cos^{-1}\left(\frac{xy}{ab} - \sqrt{\left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right)}\right) \]

### Step 2: Set the equation equal to cos(theta)
From the original equation, we can set:

\[ \cos\theta = \frac{xy}{ab} - \sqrt{\left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right)} \]

### Step 3: Rearranging the equation
Rearranging gives us:

\[ \sqrt{\left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right)} = \frac{xy}{ab} - \cos\theta \]

### Step 4: Square both sides
Squaring both sides results in:

\[ \left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right) = \left(\frac{xy}{ab} - \cos\theta\right)^2 \]

### Step 5: Expand both sides
Expanding the left side:

\[ 1 - \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{x^2y^2}{a^2b^2}\right) = \left(\frac{xy}{ab}\right)^2 - 2\frac{xy}{ab}\cos\theta + \cos^2\theta \]

### Step 6: Rearranging terms
Rearranging gives:

\[ 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2y^2}{a^2b^2} = \frac{x^2y^2}{a^2b^2} - 2\frac{xy}{ab}\cos\theta + \cos^2\theta \]

### Step 7: Canceling terms
Canceling \(\frac{x^2y^2}{a^2b^2}\) from both sides results in:

\[ 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} = -2\frac{xy}{ab}\cos\theta + \cos^2\theta \]

### Step 8: Rearranging to find the desired expression
Rearranging gives us:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + 2\frac{xy}{ab}\cos\theta - \cos^2\theta \]

### Step 9: Using the identity \(1 - \cos^2\theta = \sin^2\theta\)
Thus, we can write:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \sin^2\theta + 2\frac{xy}{ab}\cos\theta \]

### Final Result
Therefore, the final expression we are looking for is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \sin^2\theta + 2\frac{xy}{ab}\cos\theta \]