If cos−1x2+cos−1y3=θ then 9x2−12xycosθ+4y2 is equal to
36
−36sin2θ
36sin2θ
36cos2θ
We have,
cos−1x2+cos−1y3=θ⇒cos−1xy6−1−x241−y29=θ⇒xy−4−x29−y2=6cosθ⇒(xy−6cosθ)2=4−x29−y2⇒−12xycosθ+36cos2θ=36−4y2−9x2⇒9x2+4y2−12xycosθ=36sin2θ