If $\frac{x}{a}cos\theta+\frac{y}{b}sin\theta=1$ and $\frac{x}{a}sin\theta-\frac{y}{b}cos\theta=1$, prove that $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2$.

Given: $\frac{x}{a}cos\theta+\frac{y}{b}sin\theta=1$ and $\frac{x}{a}sin\theta-\frac{y}{b}cos\theta=1$

To do: To prove $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2$.

$\frac{x}{a}cos\theta+\frac{y}{b}sin\theta=1\ -----( 1)$

$\frac{x}{a}sin\theta-\frac{y}{b}cos\theta=1\ -----( 2)$

Squaring both the equations and then adding,

$\frac{x^2}{a^2}cos^2\theta+\frac{y^2}{b^2}sin^2\theta+\frac{2xy}{ab}sin\theta.cos\theta+\frac{x^2}{a^2}sin^2\theta+\frac{y^2}{b^2}cos^2\theta-\frac{2xy}{ab}sin\theta.cos\theta=2$

$\Rightarrow \frac{x^2}{a^2}( cos^2\theta+sin^2\theta)+\frac{y^2}{b^2}( sin^2\theta + cos^2\theta )=2$

$\Rightarrow \frac{x^2}{a^2}\times1+\frac{y^2}{b^2}\times1=2$       [ $\because sin^2x + cos^2x = 1$ from trigonometric identities ]

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

Hence proved.