If cos^-1(x/a)+cos^-1(y/b)=@ Then prove that x^2/a^2 …

Given, {tex}{\cos ^{ - 1}}\frac{x}{a} + {\cos ^{ - 1}}\frac{y}{b} = a{/tex}
{tex}\left[ {\because {{\cos }^{ - 1}}x + {{\cos }^{ - 1}}y} \right. \left. { = {{\cos }^{ - 1}}\left( {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right)} \right]{/tex}
{tex}\Rightarrow {\cos ^{ - 1}}\left[ {\frac{x}{a}.\frac{y}{b} - \sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} .\sqrt {1 - \frac{{{y^2}}}{{{b^2}}}} } \right] = a{/tex}
{tex}\Rightarrow\frac{{xy}}{{ab}} - \sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} \cdot \sqrt {1 - \frac{{{y^2}}}{{{b^2}}}} = \cos a{/tex}
{tex}\Rightarrow \frac{{xy}}{{ab}} - \cos a = \sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} \sqrt {1 - \frac{{{y^2}}}{{{b^2}}}}{/tex}
On squaring both side, we get
{tex}\Rightarrow{\left( {\frac{{xy}}{{ab}} - \cos a } \right)^2} = {\left( {\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} \sqrt {1 - \frac{{{y^2}}}{{{b^2}}}} } \right)^2}{/tex}
{tex}\Rightarrow \frac{{{x^2}{y^2}}}{{{a^2}{b^2}}} + {\cos ^2}a - 2.\frac{{xy}}{{ab}}.\cos a = \left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)\left( {1 - \frac{{{y^2}}}{{{b^2}}}} \right){/tex}
{tex}\Rightarrow\frac{{{x^2}{y^2}}}{{{a^2}{b^2}}} + {\cos ^2}a - 2\frac{{xy}}{{ab}}\cos a = 1 - \frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} + \frac{{{x^2}{y^2}}}{{{a^2}{b^2}}}{/tex}
{tex}\Rightarrow\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 2\frac{{xy}}{{ab}}\cos a = 1 - {\cos ^2}a{/tex}
{tex}\Rightarrow\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 2\frac{{xy}}{{ab}}\cos a = {\sin ^2}a{/tex}