Find the angle between the lines represented by `x^2+2x ysectheta+y^2=0`.

Explanation

We have,
`x^(2)+2xysectheta+y^(2)=0`
`impliesx^(2)+2x. y sec theta+y^(2)sec^(2)theta=y^(2)sec ^(2)theta-y^(2)`
`implies(x-ysectheta)^(2)=y^(2)(sec^(2)theta-1)=y^(2)tan^(2)theta`
`impliesx+y sec theta=+-y tantheta`
`impliesx+((1)/(costheta)+-(sintheta)/(costheta))y=0`
`impliesx cos theta+(1-sin theta)y=0`
`and x cos theta+(1+sin theta)y=0`
Thus, the straight lines represented by the given equation are given as
`x costheta+(1-sin theta)y=0and x costheta+(1+sintheta)y=0`
The angle between them,
`=tan^(-1)((2sqrt(sec^(2)theta0-1))/(1+1))=tan^(-1)(tantheta)=theta`