Question

eliminate theta if x=acot theta+b cosec theta, y=a cosec theta+b cot theta ​

Answer 1

$\textbf{Given:}$

$x=a\,cot\theta+b\,cosec\theta$...(1)

$y=a\,cosec\theta+b\,cot\theta$...(2)

$(1)^2-(2)^2\implies$

$x^2-y^2=(a\,cot\theta+b\,cosec\theta)^2-(a\,cosec\theta+b\,cot\theta)^2$

$x^2-y^2=(a^2cot^2\theta+b^2cosec^2\theta+2\,ab\,cot\theta\,cosec\theta)-(a^2cosec^2\theta+b^2cot^2\theta+2\,ab\,cot\theta\,cosec\theta)$

$x^2-y^2=a^2cot^2\theta+b^2cosec^2\theta+2\,ab\,cot\theta\,cosec\theta-a^2cosec^2\theta-b^2cot^2\theta-2\,ab\,cot\theta\,cosec\theta$

$x^2-y^2=a^2cot^2\theta+b^2cosec^2\theta-a^2cosec^2\theta-b^2cot^2\theta$

$x^2-y^2=a^2(cot^2\theta-cosec^2\theta)+b^2(cosec^2\theta-cot^2\theta)$

$x^2-y^2=-a^2(cosec^2\theta-cot^2\theta)+b^2(cosec^2\theta-cot^2\theta)$

$\text{Using}$

$\bf\,cosec^2A-cot^2A=1$

$\implies\,x^2-y^2=-a^2(1)+b^2(1)$

$\implies\,x^2-y^2=-(a^2-b^2)$

$\implies\boxed{\bf\,x^2-y^2+(a^2-b^2)=0}$

Answer 2

This is question 2 from Ex-22B from RS AGGARWAL (ICSE)

Hope it helps ;)