Question
If
\[ \cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha \]
Find:
\[ \frac{x^2}{a^2} – \frac{2xy}{ab}\cos\alpha + \frac{y^2}{b^2} \]
Solution
Let
\[ \cos^{-1}\left(\frac{x}{a}\right) = A,\quad \cos^{-1}\left(\frac{y}{b}\right) = B \]
Then,
\[ A + B = \alpha \]
So,
\[ \cos A = \frac{x}{a}, \quad \cos B = \frac{y}{b} \]
We use identity:
\[ \cos(A + B) = \cos A \cos B – \sin A \sin B \]
Thus,
\[ \cos \alpha = \frac{x}{a}\cdot \frac{y}{b} – \sqrt{1-\frac{x^2}{a^2}} \cdot \sqrt{1-\frac{y^2}{b^2}} \]
Rearrange and square appropriately, the standard result becomes:
\[ \frac{x^2}{a^2} – \frac{2xy}{ab}\cos\alpha + \frac{y^2}{b^2} = \sin^2\alpha \]
Final Answer:
\[ \boxed{\sin^2\alpha} \]
Key Concept
Use cosine addition identity and convert into algebraic symmetric form.