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

\[{{\sec }^{2}}(\alpha -\beta )\]

\[\cos e{{c}^{2}}(\alpha -\beta )\]

\[si{{n}^{2}}(\alpha -\beta )\]

If \[\theta \]is eliminated from the equations \[x=a\,\cos (\theta -\a

1.	If \[\theta \]is eliminated from the equations \[x=a\,\cos (\theta -\alpha )\]and \[y=b\,\cos (\theta -\beta )\], then \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{2xy}{ab}\cos (\alpha -\beta )\]is equal to
A.	\[{{\sec }^{2}}(\alpha -\beta )\]
B.	\[\cos e{{c}^{2}}(\alpha -\beta )\]
C.	\[{{\cos }^{2}}(-\beta )\]
D.	\[si{{n}^{2}}(\alpha -\beta )\]
Answer» E.

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

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