If cos−1⁡x2+cos−1⁡y3=θ, then 9x2−12xycos⁡θ+4y2=

Given, θ=cos−1⁡x2⋅y3−1−x241−y29⇒cos⁡θ=xy6−4−x29−y26⇒(xy−6cos⁡θ)2=4−x29−y2=36−9x3−4y7+x7y7⇒36cos2⁡θ−12xycos⁡θ+4y2+9x2=36∴9x2−12xycos⁡θ+4y2=36sin2⁡θ