Question

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

Answer 1

We have,

${\cos }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{y}{b}=\alpha$

We know that

${\cos }^{-1}a+{\cos }^{-1}b={\cos }^{-1}(ab-\sqrt{1-a^2}\sqrt{1-b^2})$

Therefore,

${\cos }^{-1}(\frac{x}{a}\frac{y}{b}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}})=\alpha$

$\frac{x}{a}\frac{y}{b}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}=\cos \alpha$

$\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}=\frac{x}{a}\frac{y}{b}-\cos \alpha$

On squaring both sides, we get

$(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2})=\frac{x^2}{a^2}\frac{y^2}{b^2}+{\cos }^2\alpha -2\frac{x}{a}\frac{y}{b}\cos \alpha$

$1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2}{a^2}\frac{y^2}{b^2}=\frac{x^2}{a^2}\frac{y^2}{b^2}+{\cos }^2\alpha -2\frac{x}{a}\frac{y}{b}\cos \alpha$

$1-\frac{x^2}{a^2}-\frac{y^2}{b^2}={\cos }^2\alpha -2\frac{x}{a}\frac{y}{b}\cos \alpha$

$1-{\cos }^2\alpha =\frac{x^2}{a^2}+\frac{y^2}{b^2}-2\frac{x}{a}\frac{y}{b}\cos \alpha$

${\sin }^2\alpha =\frac{x^2}{a^2}-2\frac{x}{a}\frac{y}{b}\cos \alpha +\frac{y^2}{b^2}$

Hence, proved.