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)`

Options

A. `sec^(2)(alpha-beta)`
B. `co sec^(2)(alpha-beta)`
C. `cos^(2)(-beta)`
D. `sin^(2)(alpha-beta)`

Answer

D

Explanation

`(alpha-beta)=(theta-beta)-(theta-alpha)`
`rArr cos(alpha-beta)=cos(theta-beta)cos(theta-alpha)+sin(theta+beta)sin(theta-alpha)`
`=(y)/(b)xx(x)/(a)+sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))`
`rArr[(xy)/(ab)-cos(alpha-beta)]^(2)=(1-(x^(2))/(a^(2)))(1-(y^(2))/(b^(2)))`
or `(x^(2)y^(2))/(a^(2)b^(2))+cos^(2)(alpha-beta)-(2xy)/(ab)cos(alpha-beta)`
`1-(y^(2)/(b^(2))-(x^(2))/(a^(2))+(x^(2)y^(2))/(a^(2)b^(2))`
or `(x^(2))/(a^(2))+(y^(2))/(b^(2))-(2xy)/(ab)cos(alpha-beta)=sin^(2)(alpha-beta)`