If `p` and `q` are the lengths of perpendicular from the origin to the line `xcos(theta)-ysin(theta)=kcos(2theta)` and` xsec(theta)+ycosec(theta)=k` respectively , then prove that `p^2+4q^2=k^2`

Explanation

Length of perpendicular from origin to the line `x sec theta+y cosec theta-k=0` is
`q=|(0+0-k)/(sqrt(sec^(2)theta+cosec^(2)theta))|=(k)/(sqrt((1)/(cos^(2)theta)+(1)/(sin^(2)theta)))`
`q=(|k sin theta cos theta|)/(sqrt(sin^(2)theta+cos^(2)theta))=|k sinthetacostheta|`
`implies 2q=|k (2sinthetacostheta)|=|k sin 2theta|`
`implies 4q^(2)=k^(2)sin^(2)2theta` .......`(1)`
and length of perpendicular from origin to line `x cos theta-y sin theta-k cos 2theta=0` is,
`p=|(0+0-kcos2theta)/(sqrt(cos^(2)theta+sin^(2)theta))|=|k cos 2theta|`
`implies p^(2)=k^(2)cos^(2)2theta` ..........`(2)`
Now, adding equations `(1)` and `(2)`
4q^(2)+p^(2)=k^(2)sin^(2)2theta+k^(2)cos^(2)2theta`
`=k^(2)(sin^(2)2theta+cos^(2)2theta)=k^(2)`