Given equation, x‌cos‌θ−y‌sin‌θ=2‌cos‌2‌θ ⇒y=x‌cot‌θ−
2‌cos‌2‌θ
sin‌θ
∴ Slope of line =cot‌θ Then, slope of line perpendicular to given line =
−1
cot‌θ
=−tan‌θ Let equation of line which is perpendicular to x‌cos‌θ−y‌sin‌θ=2‌cos‌2‌θ, is y=mx+C ...(i) y=(−tan‌θ)x+C ...(ii) [∵ its slope is −tan‌θ] Since, Eq. (i) passes through (2cos3θ,2sin3θ) From, Eq. (ii), 2sin3θ=−tan‌θ.2cos3θ+C ⇒C=2sin3θ+2‌sin‌θ‌cos2‌θ ∴ Eq. (ii), becomes y=(−tan‌θ)x+2sin3θ+2‌sin‌θ‌cos2‌θ y=−
x‌sin‌θ
cos‌θ
+2sin3θ+2‌sin‌θ‌cos2‌θ ⇒y‌cosec‌θ=−x‌sec‌θ+2(sin2θ+cos2θ) ⇒y‌cosec‌θ+x‌sec‌θ=2 i.e. x‌sec‌θ+y‌cosec‌θ=2