Question

If $\theta$ is eliminated from the equations $x=acos(\theta -\alpha )$ and $y=bcos(\theta -\beta )$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}cos(\alpha -\beta )$ is equal to

• A. $se{c}^2(\alpha -\beta )$
• B. $cose{c}^2(\alpha -\beta )$
• C. $co{s}^2(-\beta )$
• D. $si{n}^2(\alpha -\beta )$

Answer 1

Answer: (D)

$(\alpha -\beta )=(\theta -\beta )-(\theta -\alpha )$
$\Rightarrow cos(\alpha -\beta )=cos(\theta -\beta )cos(\theta -\alpha )$
$+sin(\theta -\beta )sin(\theta -\alpha )$
$\Rightarrow {\left[\frac{xy}{ab}-cos\left(\alpha -\beta \right)\right]}^2=\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)$
or $\frac{x^2{y}^2}{a^2{b}^2}+co{s}^2(\alpha -\beta )-\frac{2xy}{ab}cos(\alpha -\beta )$
$=1-\frac{y^2}{b^2}-\frac{x^2}{a^2}+\frac{x^2{y}^2}{a^2{b}^2}$
or $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}cos(\alpha -\beta )=si{n}^2(\alpha -\beta )$