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If `theta`is eliminated from the equations `x=a"cos"(theta-alpha)`and `y=bcos(theta-beta),`then`sec^2(alpha-beta)`(b) `cos e c^2(alpha-beta)``cos^2(-beta)`(d)`sin^2(alpha-beta)`A. `sec^(2)(alpha-beta)`B. `co sec^(2)(alpha-beta)`C. `cos^(2)(-beta)`D. `sin^(2)(alpha-beta)` |
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Answer» Correct Answer - D `(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)` |
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