If `theta` is eliminated from the equations `x=a"cos"(theta-alpha)` and `y=bcos(theta-beta),` then `(x^2/a^2)+(y^2/b^2)-(2xy)/(ab)cos(alpha-beta)` is equal to (a)`sec^2(alpha-beta)` (b) `cos e c^2(alpha-beta)` (c)`cos^2(alpha-beta)` (d) `sin^2(alpha-beta)`

Explanation

`cos(theta-alpha)=x/a`and`cos(theta-beta)=y/b`

`sin^2(theta-alpha)=1-cos^2(theta-alpha)`

`=(1-x^2/a^2)`

`sin(theta-alpha)=sqrt(1-x^2/a^2)`

`(alpha-beta)=(theta-beta)-(theta-alpha)`

`cos(alpha-beta)=cos[(theta-beta)-(theta-alpha)]`

`=cos(theta-beta)cos(theta-alpha)+sin(theta-beta)sin(theta-alpha)`

`=x/a*y/b+sqrt(1-x^2/a^2)*sqrt(1-y^2/b^2)`

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