If `cos^(-1)x/a+cos^(-1)y/b=alpha,` prove that `(x^2)/(a^2)-2(x y)/(a b)cosalpha+(y^2)/(b^2)=sin^2alpha`

Answer

Proved

Explanation

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


`\quad[ \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}(x y-\sqrt{(1-x^{2})(1-y^{2})})]`


` \cos ^{-1}[\frac{x}{a} \cdot \frac{y}{b}-\sqrt{.(1-\frac{x^{2}}{a^{2}})(1-\frac{y^{2}}{b^{2}})}]=\alpha`


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


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


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


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


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


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