# If $\cos^{-1}\frac{x}{a}+\cos^{-1}\frac{y}{b}=\alpha,$prove that $\frac{x^2}{a^2}-\frac{2xy}{ab}\cos \alpha+\frac{y^2}{b^2}=\sin^2\alpha.$

Toolbox:
• $$cos^{-1}x+cos^{-1}y=cos^{-1} \bigg[ xy+\sqrt{1-x^2}. \sqrt{1-y^2} \bigg]$$
• $$cos^2\alpha=1-sin^\alpha$$
Given $$cos^{-1}\large\frac{x}{a}+cos^{-1}\large\frac{y}{b}=\alpha$$

$$Taking\:\large\frac{x}{a}\:in\:the\:place\:of\:x\:and\:$$
$$\large\frac{y}{a}\:in\:the\:place\:of\:y\:in\:the\:above\:formula$$

$$\Rightarrow cos^{-1} \bigg[ \large\frac{xy}{ab}-\sqrt{1-\large\frac{x^2}{a^2}}\sqrt{1-\large\frac{y^2}{b^2}} \bigg] = \alpha$$
$$\Rightarrow cos\alpha = \large\frac{xy}{ab}-\sqrt{1-\large\frac{x^2}{a^2}}\sqrt{1-\large\frac{y^2}{b^2}}$$
$$\Rightarrow cos\alpha - \large\frac{xy}{ab}=-\sqrt{1-\large\frac{x^2}{a^2}}\sqrt{1-\large\frac{y^2}{b^2}}$$

squaring both sides and write $$cos^2\alpha = 1-sin^2\alpha$$
$$\Rightarrow\:cos^2\alpha+(\frac{xy}{ab})^2-2\large\frac{xy}{ab}cos\alpha=$$
$$\bigg(1-\large\frac{x^2}{a^2}\bigg)\bigg(1-\large\frac{y^2}{b^2}\bigg)=1+(\large\frac{xy}{ab})^2-\large\frac{x^2}{a^2}-\large\frac{y^2}{b^2}$$

$$\Rightarrow\:\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}-2\large\frac{xy}{ab}=sin^2\alpha$$

answered Feb 20, 2013
edited Mar 19, 2013