If (x/a sin A - y/b …

Given: {tex}\frac { x } { a } \cos \theta + \frac { y } { b } \sin \theta = 1 \text { and } \frac { x } { a } \sin \theta - \frac { y } { b } \cos \theta = 1 , {/tex}
To prove:  {tex}\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 2{/tex}
Now, {tex}{\left( {\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta } \right)^2} + {\left( {\frac{x}{a}\sin \theta - \frac{y}{b}\cos \theta } \right)^2} = {(1)^2} + {(1)^2}{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + 2\frac{x}{a}\cos \theta \frac{y}{b}\sin \theta + \frac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta - 2\frac{x}{a}\sin \theta \frac{y}{b}\cos \theta = 1 + 1{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \frac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + \frac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta = 2{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + \frac{{{y^2}}}{{{b^2}}}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 2{/tex}
{tex} = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 2{/tex} {tex}\left[ {\because {{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right]{/tex}
Hence proved.