Question

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

• A. sec²(α-β)
• B. cosec²(α-β)
• C. cos²(-β)
• D. sin² (α-β)

Answer 1

The correct option is D

sin² (α-β)

Calculate the value of equation.

It is given that the equations are,

$x=a\cos (\theta -\alpha )$

We can rewrite it as,

$\cos (\theta -\alpha )=\frac{x}{a}$ ------- $\left(1\right)$

$y=b\cos (\theta -\beta )$

$\cos (\theta -\beta )=\frac{y}{b}$ ------- $\left(2\right)$

From the above value $\cos (\theta -\alpha )$ and $\cos (\theta -\beta )$, we can get the value of $\sin (\theta -\alpha )$ and $\sin (\theta -\beta )$.

Use Pythagorean identity of trigonometric function.

$\Rightarrow$ ${\sin }^2\left(\theta -\alpha \right)=1-{\cos }^2\left(\theta -\alpha \right)$

$\Rightarrow$ ${\sin }^2\left(\theta -\alpha \right)=1-\frac{x^2}{a^2}$

$\Rightarrow$ $\sin \left(\theta -\alpha \right)=\sqrt{1-\frac{x^2}{a^2}}$ ------- $\left(3\right)$

Similarly,

$\Rightarrow$ $\sin \left(\theta -\beta \right)=\sqrt{1-\frac{y^2}{b^2}}$ --------- $\left(4\right)$

$\Rightarrow$ $\alpha -\beta =\left(\theta -\beta \right)-\left(\theta -\alpha \right)$

Taking $\cos$ both sides.

$\Rightarrow$ $\cos \left(\alpha -\beta \right)=\cos \left(\left(\theta -\beta \right)-\left(\theta -\alpha \right)\right)$

Use the formula $\cos \left(a-b\right)$.

As we know that the formula for $\cos \left(a-b\right)$ is,

$\mathbf{cos}\left(a-b\right)\mathbf{=}\mathbf{cos}\left(\mathit{a}\right)\mathbf{\times }\mathbf{cos}\left(\mathit{b}\right)\mathbf{+}\mathbf{sin}\left(\mathit{a}\right)\mathbf{\times }\mathbf{sin}\left(\mathit{b}\right)$

So,

$\Rightarrow$ $\cos \left(\alpha -\beta \right)=\cos \left(\theta -\beta \right)·\cos \left(\theta -\alpha \right)+\sin \left(\theta -\beta \right)·\sin \left(\theta -\alpha \right)$

Put all the values from equation $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$.

$\Rightarrow$ $\cos \left(\alpha -\beta \right)=\frac{x}{a}\times \frac{y}{b}+\sqrt{1-\frac{x^2}{a^2}}·\sqrt{1-\frac{y^2}{b^2}}$

$\Rightarrow$ $-\sqrt{1-\frac{x^2}{a^2}}·\sqrt{1-\frac{y^2}{b^2}}=\left[\frac{xy}{ab}-\cos \left(\alpha -\beta \right)\right]$

Squaring both sides, we get,

$\Rightarrow$ $\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)={\left[\frac{x^{2}y}{ab}-\cos \left(\alpha -\beta \right)\right]}^2$

$\Rightarrow$ $1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2{y}^2}{a^2{b}^2}=\frac{x^2{y}^2}{a^2{b}^2}+{\cos }^2\left(\alpha -\beta \right)-\frac{2xy}{ab}\cos \left(\alpha -\beta \right)$

Cancel common factor. We get,

$\Rightarrow$ $1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\cancel{\frac{x^2{y}^2}{a^2{b}^2}}=\cancel{\frac{x^2{y}^2}{a^2{b}^2}}+{\cos }^2\left(\alpha -\beta \right)-\frac{2xy}{ab}\cos \left(\alpha -\beta \right)$

$\Rightarrow$ $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos \left(\alpha -\beta \right)=1-{\cos }^2\left(\alpha -\beta \right)$

$\Rightarrow$ $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos \left(\alpha -\beta \right)=1-{\sin }^2\left(\alpha -\beta \right)$

Hence, option $\mathrm{D}$ is correct answer.