If cos−1x−cos−1y3=α\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alphacos−1x−cos−13y=α, where −1≤x≤1,−3≤y≤3,x≤y3-1 \leq x \leq 1,-3 \leq y \leq 3, x \leq \frac{y}{3}−1≤x≤1,−3≤y≤3,x≤3y, then for all x,y9x2−6xycosα+y2x, y 9 x^{2}-6 x y \cos \alpha+y^{2}x,y9x2−6xycosα+y2 is equal to
sin2α\sin ^{2} \alphasin2α
3sin2α3 \sin ^{2} \alpha3sin2α
9sin2α9 \sin ^{2} \alpha9sin2α
49sin2α\frac{4}{9} \sin ^{2} \alpha94sin2α
Correct Answer: C — 9sin2α9 \sin ^{2} \alpha9sin2α
Explanation:
cos−1a−cos−1 b=cos−1(ab+1−a2⋅1−b2) ∴cos−1x−cos−1y3 =cos−1(xy3+1−x2⋅1−y29)=a ∴xy3+1−x2⋅9−y23=cosα xy+1−x2⋅9−y2=3cosα xy−3cosα=−1−x2⋅9−y2\begin{aligned} & \cos ^{-1} \mathrm{a}-\cos ^{-1} \mathrm{~b}=\cos ^{-1}\left(a b+\sqrt{1-\mathrm{a}^{2}} \cdot \sqrt{1-\mathrm{b}^{2}}\right) \\\ \therefore & \cos ^{-1} x-\cos ^{-1} \frac{y}{3} \\\ = & \cos ^{-1}\left(\frac{x y}{3}+\sqrt{1-x^{2}} \cdot \sqrt{1-\frac{y^{2}}{9}}\right)=a \\\ \therefore & \frac{x y}{3}+\frac{\sqrt{1-x^{2}} \cdot \sqrt{9-y^{2}}}{3}=\cos \alpha \\\ & x y+\sqrt{1-x^{2}} \cdot \sqrt{9-y^{2}}=3 \cos \alpha \\\ & x y-3 \cos \alpha=-\sqrt{1-x^{2}} \cdot \sqrt{9-y^{2}} \end{aligned} ∴ = ∴ cos−1a−cos−1 b=cos−1(ab+1−a2⋅1−b2)cos−1x−cos−13ycos−1(3xy+1−x2⋅1−9y2)=a3xy+31−x2⋅9−y2=cosαxy+1−x2⋅9−y2=3cosαxy−3cosα=−1−x2⋅9−y2 squaring on both sides, we get x2y2−6xycosα+9cos2α=(1−x2)(9−y2)x^{2} y^{2}-6 x y \cos \alpha+9 \cos ^{2} \alpha=\left(1-x^{2}\right)\left(9-y^{2}\right)x2y2−6xycosα+9cos2α=(1−x2)(9−y2) x2y2−6xycosα+9cos2α=9−y2−9x2+x2y2x^{2} y^{2}-6 x y \cos \alpha+9 \cos ^{2} \alpha=9-y^{2}-9 x^{2}+x^{2} y^{2}x2y2−6xycosα+9cos2α=9−y2−9x2+x2y2 i.e., 9x2−6xycosα+y2=9sin2α9 x^{2}-6 x y \cos \alpha+y^{2}=9 \sin ^{2} \alpha9x2−6xycosα+y2=9sin2α
