If cos^{-1}left(frac{x}{a}right) + cos^{-1}le | vedclass

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Question

(A) We are given $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$. Using the identity $\cos^{-1} A + \cos^{-1} B = \co

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$\sin^2 \alpha$

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(A) We are given $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$. Using the identity $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$,we get: $\cos^{-1} \left[ \frac{x}{a} \cdot \frac{y}{b} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} \right] = \alpha$. Taking $\cos$ on both sides: $\frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} = \cos \alpha$. Rearranging the terms: $\frac{xy}{ab} - \cos \alpha = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$. Squaring both sides: $\left( \frac{xy}{ab} - \cos \alpha \right)^2 = \left( 1 - \frac{x^2}{a^2} \right) \left( 1 - \frac{y^2}{b^2} \right)$. $\frac{x^2 y^2}{a^2 b^2} - \frac{2xy}{ab} \cos \alpha + \cos^2 \alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2 y^2}{a^2 b^2}$. Canceling $\frac{x^2 y^2}{a^2 b^2}$ from both sides: $-\frac{2xy}{ab} \cos \alpha + \cos^2 \alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}$. Rearranging to isolate the required expression: $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = 1 - \cos^2 \alpha = \sin^2 \alpha$.

If $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$,then $\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = $

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