Prove the following trigonometric identities.

Given that,

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

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

We have to prove $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2$

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

Squaring and then adding the above two equations, we have

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

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

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

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