If $frac{x}{a}-frac{y}{b} an heta=1$ and $frac{x}{a} an heta+frac{y}{b}=1$, then the value of $frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}$ is

$frac{x}{a}-frac{y}{b} an heta=1$

$frac{x}{a} an heta+frac{y}{b}=1$

On squaring equations (i) and (ii) and adding,

$frac{x^{2}}{a^{2}}-frac{2 x y}{a b} an heta+frac{y^{2}}{b^{2}} an ^{2} heta+frac{x^{2}}{a^{2}} an ^{2} heta$

$+frac{2 x y}{a b} an heta+frac{y^{2}}{b^{2}}=1+1$

$Rightarrow frac{x^{2}}{a^{2}}+frac{x^{2}}{a^{2}} an ^{2} heta+frac{y^{2}}{b^{2}} an ^{2} heta+frac{y^{2}}{b^{2}}=2$

$Rightarrow frac{x^{2}}{a^{2}}left(1+ an ^{2} heta ight)+frac{y^{2}}{b^{2}}left( an ^{2} heta+1 ight)=2$

$Rightarrow frac{x^{2}}{a^{2}} sec ^{2} heta+frac{y^{2}}{b^{2}} sec ^{2} heta=2$

$Rightarrow frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=frac{2}{sec ^{2} heta}=2 cos ^{2} heta$