Trig1.6

From Exampleproblems

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

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

Squaring and adding the two equations,we get

$[\frac{x}{a}\sin\theta+\frac{y}{b}\cos\theta]^2+[\frac{x}{a}\cos\theta-\frac{y}{b}\sin\theta]^2=1+1\,$

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